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Chap. ....SB.S_S.iL 



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UNITED STATES OF AMERICA. 



SELF-INSTRUCTOR 



LUMBER SURVEYING-, 



FOE THE USE OF 



LUMBER MANUFACTURERS, SURVEYORS, 
AND TEACHERS. 



by / 
OHAELES KINSLEY, 

PRACTICAL SURVEYOR AND TEACHER OF SURVEYING. 



ASSIGNOR, 
JAMES KINSLEY. 




PUBLISHED BY THE AUTHOR. 

Calais, Me., and St. Stephen, N. B. 

1870. 



Entered according to Act of Congress, in the year 1870, 

Br Charles Kinsley, 

In the Office of the Librarian of Congress, at Washington. 



RIVERSIDE, CAMBRIDGE: 

ELECTROTYPED AND PRINTED BY 

H. 0. HOUGHTON AND COMPANY. 



^ 



PREFACE. 



This work combines the theoretical and practical 
parts of surveying, in such a manner as to enable the 
energetic and uninitiated student who applies himself 
to the study of this useful and interesting science for a 
short time, to survey all kinds of lumber with accu- 
racy and expertness. It contains tables for measuring 
boards, plank, deal, and timber by board measure, 
by which the Surveyor can dispense with the use of 
the Board Rule. It contains the rules generally 
adopted by Surveyors, and also a more concise rule 
than that in general use : for plank, deal, and timber, 
this rule alone is worth more than the price of the 
book to any Surveyor, as it requires less mental calcu- 
lation than by the other rules, enabling him to sur- 
vey faster and with less trouble than he could other- 
wise do. It contains tables for inch, inch and a 
quarter, and inch and a half boards for battens and 
joist. It also contains rules and tables for surveying 
logs by board and cubic measure, and rules for ton tim- 
ber. It also contains tables showing the number 
of feet in length, of any dimension, which will make 
1,000 feet board measure or 1,000 feet cubic measure ; 



IV PREFACE. 

a new method of finding the solid contents of timber ; 
a rule for finding what a round log will square, by hav- 
ing the circumference or diameter given, or in other 
words, to find the inscribed square ; how to make out 
specifications, survey bills, etc. ; rule for measuring 
tapering timber ; table of quarter-girts for logs ; rule 
for finding how much in length, of any dimension, 
which will make a solid foot, or any other desired 
quantity; table showing the weight of twenty-five 
kinds of wood, with a rule for finding the weight of 
the same from the contents ; the English and Amer- 
ican Government rules for finding the tonnage of ves- 
sels, and rules for gauging and ullaging casks. It also 
contains a correct and extensive interest table. 



TABLE OF CONTESTS. 



Page 

Rule for measuring Rectangular Boards 7 

Exercise on the Rule 7 

To find the Contents of a Triangular Board 8 

Exercise on the Rule 8 

To find the Solidity of a Sphere or Globe 9 

To find the Contents of a Triangular Solid 9 

To find the Superficial Contents of a Globe 9 

To find the Contents of a Circular Board 9 

Rule for finding the Contents of a Circle 9 

Rule for finding the Diameter or Circumference of a Circle . . 9 
Table for measuring Inch Boards, without a Board Rule, from Two 

Inches to Thirty-six Inches wide 10 

Exercise on the Table 10 

Table for measuring Inch-and-a-Quarter Boards without a Board Rule 11 

Exercise on the Table 11 

Table for Inch-and-a-Half Boards without using the Board Rule . . 12 

Exercise on the Rule 12 

Table for Plank from Two to Thirty Inches wide . . . . .13 

Exercise on the Table 13 

Table for Three-inch Deals from Three to Twenty-four Inches wide . 14 

Table for Four inch Deals from Four to Twelve Inches wide . . 15 

Table for Five-inch Timber from Five to Twelve Inches wide . . .16 

Table for Six-inch Timber from Six to Twelve Inches wide . . 16 

Table for Seven -inch Timber from Seven to Twelve Inches wide . . 17 

Table for Eight-inch Timber from Eight to Twelve Inches wide . 17 

Table for Nine-inch Timber from Nine to Twelve Inches wide . . 17 

Table for Ten-inch Timber from Ten to Twelve Inches wide . . 17 

Table for Eleven-inch Timber from Eleven to Twelve Inches wide . 17 

Table for Twelve-inch Timber from Twelve to Twenty Inches wide . 17 
Plan of drawing a Plank Shingle, with Directions for dotting Plank, 

Deal, &c 18,19 

Rule for Plank Specifications 20 

How to keep a Joist or Scantling Shingle 22 

New York Deals, Three-inch 23 

Specification Rule for Three-inch 24 

Specification Rules for Four-inch 25, 26 

Five-inch Timber Shingle, with Rule for Specification . . . 28-30 



VI TABLE OF CONTENTS. 

Rule for Six-inch Specification 31 

Rule for Seven-inch Specification 33 

Rule for Eight-inch Specification 35 

Rule for Nine-inch Specification ........ 37 

Rule for Ten-inch Specification 40 

Rule for Eleven-inch Specification . 41 

Rule for Twelve inch Specification 42 

Rule for finding the Contents of Battens or Two-and-a-Half-inch Stuff 44 

Specification of Batten Shingle and Rule 45 

Random Shingle, Contents given in the Columns 45 

Random Shingle, Running Lengths in the Columns .... 46 
Table showing the Number of Feet in Length, of all Dimensions, that 

will make 1,000 feet of Board Measure 47 

Table showing the Number of Feet in Length, of all Dimensions, from 
Five Inches by Five Inches to Twenty-two Inches by Twenty- 
Four Inches that will make 1,000 Cubic feet, with Rules showing 

how both Tables are computed 48 

New Rules for finding the Contents in Cubic Feet of Timber, from Five 

by Five up . . 49 

Second Method of making out Specification, and Rule . . .51, 52 

Specification of Philadelphia Deal, and Rule 53, 54 

How to use the Board R'ule, with Exercise 54 

Rule for measuring Logs, with Example 55 

To find the Largest Square Piece of Timber that can be sawed from 

a Round Log, by having the Circumference or Diameter given . 56/'"' 

Form of Bills of Lading 57, 58 

Surveyor's Bills and Receipts 59 

New Rules for finding the Superficial Contents of Plank, Deal, Joist, 

Battens, and Timber 59 

Given the Dimensions of the End of a Plank to find what Length of 

it will make a Foot 61 

To find the Solid Contents of a Piece of Tapering Timber ... 61 
When a Board or Plank is wider at one End than the other, to find 

what Length of it will make a Foot, or any Desired Quantity . 62 
To find how much in Length will make a Solid Foot, or any other De- 
sired Quantity, of Squared Timber of Equal Dimensions from End 

to End 62 

Table for measuring Round Timber by the Quarter-girt Areas . . 63 
Table and Rule for finding the Weight of Timber from a Survey of its 

Contents 64 

English Government Rule for finding the Tonnage of Vessels . . 65 

United States Government rule for finding the Tonnage of Vessels . 66 

Gauging of Casks 67 . 

Ullaging of Casks 68 

Questions for Exercise 69 

Log Rule for Round Timber 72-76 

Directions for using the Log Rule ....... 77 

Interest Table and Rule 77-79 



SELF-INSTRUCTOR 

ON 

LUMBEE SURVEYING. 



Rule for measuring Rectangular Boards. 

Multiply the length in feet by the width in inches, and 
divide the product by 12, to find the contents in superficial 
feet. Or multiply the length in inches by the width in 
inches, and divide by 144, the number of inches in a square 
foot, for the contents in superficial feet. 

P. S. — A Rectangle is a plain figure bounded by four 
straight lines, which are equal and parallel, and whose 
angles are right angles, 



as B. 



B. 



QUESTIONS FOR EXERCISE. 

1. What are the contents in feet of a rectangular board 
30 feet long and 20 inches wide ? Ans. 50 feet. 

2. How many feet in a board 26 feet 6 inches long, 12 
inches in width ? Ans. 26 J feet. 

3. What will be the cost of a walnut board 32 feet long 
and 16 inches wide, at 8 cents per square foot. Ans. $3.41. 

4. "What are the contents of a board 22 feet 8 inches long, 
and 1 foot 9 inches in width ? Ans. 39 feet 8 inches. 

When a Board is wider at one End than at the other. 

Rule. — Add the width of both ends together, and take 
half the sum for a mean width, and multiply the width thus 
found by the length, for the contents ; or take the width in 



8 



SELF-INSTRUCTOR 



the middle of the board and multiply by the length, for the 
contents. 

EXAMPLE. 

1. What are the contents of a board 14 inches at one end 
and 20 inches at the other, and 24 feet in length. 

Ans. 34 feet. 
14 _L. 20 = 34 -;- 2 = 17, mean width in inches, which 
multiplied by the length, 24 feet = 408; 408 -r- 12 = 34 
feet = contents. 

2. What are the contents of a board 26 feet long, which 
measures i6 inches in the middle ? Ans. 34 feet 8 inches. 

26 feet X 16 = 416 ; 416 ~ 12 = 34 feet 8 inches = 
contents. . 

To find the Contents of a Triangular Board. 
Rule. — Multiply the length in feet by the width in 
inches, and take half the sum for the contents in inches, 
which being divided by 12 will give the contents in feet of 
board measure. 

EXAMPLE. 

1. What are the contents of the board 
ABC, whose base B C is 26 inches, 
and perpendicular height A D is 18 
feet. Ans. 19 feet 6 inches. 

18 X 26 = 468 -r- \ = 234 -f- 12 
= 19 feet 6 inches. b 

2. What are the contents of the trian- 
gular board ABC, whose base B C is 
2 feet 6 inches, and perpendicular A C, 24 
feet. Ans. 30 feet. 

24 feet X 2J = 60 feet ; 60 feet + 2 
= 30 feet. Or— - 

2 feet 6 inches = 30 inches ; 30 inches 
X 24 feet = 720 inches; 720 -f- 2 = 360 
inches = contents; 360 -7- 12 = 30 feets 
= contents in feet. 





ON LUMBER SURVEYING. 



9 



The contents of a triangular solid can be found in the 
same manner by the foregoing rule, by multiplying the con- 
tents thus found by the thickness of the solid. 

How many feet of boards in a triangular 
piece of timber, ABC, whose length A 
B is 24 feet, breadth B C 18 inches, and 
thickness C E 2 feet 6 inches ? 

24 feet X 18 inches = 432 ; 432 -^ 2 
= 216 inches; 216 inches -f- 12 = 18 
feet = contents of superficial triangle A B 
C, which being multiplied by the thickness, 
C E, 2 feet 6 inches, will give the contents 
of the solid triangle A B C D E F, 18 feet X 2 J feet = 
Ans. 45 cubic feet, or 540 board measure. 




For Measurement of a Globe. 

Rule. — To find the solidity of a globe, cube the diame- 
ter, and multiply the product by 5,236 ; and to find the sur- 
face of a globe, multiply the diameter 
by the circumference. To find the cir- 
cumference by having the diameter 
given, say as 7 is to 22, so is the diame- 
ter to the circumference, or as 22 is to 7, 
so is the circumference to the diameter. 

To find the Contents of a Circle. 

Rule 1. — Multiply half the circum- 
ference by half the diameter, for the contents. 

Rule 2. — Square the diameter, and multiply it by .7854 
for the contents, or square the circumference, and multiply it 
by .07958 for the contents. 

P. S. — The square of a number is found by multiplying 
the number by itself. 




10 



SELF-INSTKUCTOR 



Table for measuring Inch Boards without a Rule, from 
2 Inches to 36 Inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches, Feet. 


Inches. Feet. 


2X1=| 
3X1 = | 


nXi = H 


20 X 1 = if 


29 X l = 2fV 


12x1 = 1 


2lXl = l| 


30 X 1 = 2 J 


4X1=| 


13X1 = 1 T V 


22Xl = lf 


31 X 1 = 2 T 7 2 


6X1 = A 


i4Xi = ii 


23 X 1 = Hh 


32 X l = 2f 


exi=i 


i5Xi = ii 


24 X 1 = 2 


33 X l=2j 


7X1 = T 7 2 


16Xl = lJ 


25 X 1 = 2 T V 


34X1=2| 


8X1 = | 


17 X 1 = IfV 


26 X 1 = H 


35X1 = 2H 


9X1 = | 


18 x i = ii 


27X1=2^ 


36 X 1 = 3 


10X1 = 1 


19 X 1 = 1 T 7 2 


28 X 1 = 2 J 





In order to survey boards by the Table of Board Meas- 
ure, the Surveyor must commit the table to memory, and 
by a little practice, he will become expert at surveying by 
this method. 



Questions for JExercise done by the Table of Board 
Measure. 

1. What are the contents of a board 24 feet long and 18 
inches wide? Ans. 24 X 1^ = 36 feet. 

2. How many feet in a board 32 feet long and 17 inches 
wide ? Ans. 45 J feet. 

By the table, 17 inches wide is lj\ the length, for the 
contents ; therefore 32 feet X 1-& = 45^ feet. 

3. What are the contents of a board 21 feet 6 inches long 
and 6 inches wide ? Ans. 10 feet 9 inches. 

By the table, 6 inches wide is half the length, for the con- 
tents; therefore 21 feet 6 inches — 2 = 10 feet 9 inches == 
contents. 

4. Required the contents of a board 36 feet long and 3 
inches wide? Ans. 36 -4- 4 = 9 feet. 

5. Find the contents of a board 24 feet 8 inches long and 
14 inches wide ? 

Ans. 24 feet 8 inches X H = 28 feet 9 inches 4". 



ON LUMBER SURVEYING. 



11 



6. Required the contents of a board 27 feet long and 30 
inches wide ? Arts. 67 J feet. 

7. What is the value of a walnut board 23 feet 6 inches 
long, and 36 inches wide, @ 12J- cents per square foot? 

Ans. $8.81J. 

8. Required the contents of a board 16 feet long and 27 
inches wide ? Ans. 36 feet. 

9. How many feet in a board 38 feet long and 28 inches 
wide ? Ans. 88 feet 8 inches. 

10. Required the contents of a board 16 feet long and 19 
inches in width ? Ans. 25 feet 4 inches. 



Table for Inch-and-a- Quarter Boards, from 2 Inches to 
36 Inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


2xii=A 


i4Xii=Hi 


26Xli = 2if 


BXii=tV 


15Xli = l T 9 ^ 


27Xli = 2|f 


4Xii=A 


i6xii=if 


28Xli = 2H 


5Xl± = i* 


i7xii==ifi 


29X1± = 3A 


6Xii=f 


18Xli = lJ 


30Xli = 3l 


7X1± = H 


19Xli = l|i 


sixii=s« 


8Xii=f 


20Xli = 2 T L 


32Xli = 3i 


9Xli = |f 


2ixii=aft 


33 X li ~ 3 T V 


ioxii=i^ 


22 X l£ = 2& 


34Xli = 3i| 


iiXiJ = iA 


23 X 1± = 2|! 


35Xli = 3fi 


i2Xii=ii 


24Xl± = 2j . 


36Xli = 3| 


13X1± = 1« 


25 X li = 2|| 





Examples of \\-inch Board Measure done by the Table. 

1. What are the contents of a board 1J inches thick, 32 
inches wide, and 30 feet long ? Ans. 100 feet. 

By the table 32 inches is 3^ times the length ; for the con- 
tents, therefore, 30 feet X H = 100 feet. 

2. What are the contents of a board 1 J inches by 18 inches, 
and 36 feet in length? Ans. 67 feet 6 inches. 



12 



SELF-INSTRUCTOR 



3. Required the contents of a board 1J inches by 24 
inches, and 32 feet 8 inches in length? 

Ans. 81 feet 8 inches. 

4. How many feet in a 1^-inch board 16 inches wide 
and 24 feet long ? Ans. 40 feet. 

5. What will be the cost of a piece of mahogany 1 J inches 
by 12 inches, and 36 feet long, @ 6 cents per foot ? 

Ans. $2.70. 

Table for One-and-a- Half -inch Boards, from 2 to 24 
Inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


2Xl* = i 


sx4=i 


1.4X4= If 


20x4=4 

2lX4=2f 


3X4 = 1 


9X4 = 4 


15 x 4=4 


4xii=i 
5X4=* 


10x4=4 


16X4 = 2 


22X4 = 2} 


nxii=i| 


17X4 = 21 


23X4 = 2| 
24 X 4 = 3* 


ex4=f 


12 x 4=4 


isx4 = 2 i 


7X4 = | 


13 x 4=4 


i9x4= 2 l 





1. What are the contents of a lj-inch board 32 feet long 
and 24 inches wide ? Ans. 32 feet X 3 feet = 96 feet. 

2. Required the contents of a lj-inch board 18 feet long 
and 18 inches wide ? Ans. 40 J- feet. 

3. Find the contents of a board \\ X 10 inches and 28 
feet 8 inches in length ? Ans. 35 feet 10 inches, 

By the table lj X 10 is 1J the length, for the contents. 
28 feet 8 inches X 1J = 35 feet 10 inches. 

4. What are the contents of a board 24 feet long, 20 
inches wide, and 1-J inches thick? Ans. 60 feet. 

5. Required the contents of a board 16 inches wide, 1J- 
inches thick, and 27 feet long. Ans. 54 feet. 

6. What is the value of a board 17 inches wide, and 4 
inches thick, and 20 feet long, at 6 cents per foot ? 

Ans. $2.55. 

* Equal three times the length, for contents. 



ON LUMBER SURVEYING. 



13 



Table for Two-inch or Plank, from 2 to 30 Inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


2X2 = i 

2X3 = | 


2X10 = 1$ 


2Xl7 = 2f 


2 X 24 = 4 


2XH = lf 


2 X 18 = 3 


2 X 25 = 4^ 


2X4 = f 


2 X 12 = 2 


2Xl9 = 3j 


2 X 26 = 4 j- 


2X5 = 1 


2X13 = 21- 


2X20 = 3j 


2 X 27 = 4^ 


2X6=1 


2X14 = 2j 
2X15 = 21? 


2X21 = 3^ 


2X28 = 4§ 


2X8 = lJ 
2X9 = lJ 


2X22 = 3f 


2X29 = 4f 


2X16 = 2§ 


2X23 = 3f 


2 X 30 = 5 









EXERCISE. 

1. Required the contents of a plank 18 feet long and 15 
inches in width ? Ans. 45 feet. 

By the table 15 inches wide is 2 \ times the length, for the 
contents in feet of board measure; therefore 18 feet X 2^ 
= 45 feet. 

2. Required the contents of a plank 36 feet long and 12 
inches wide at one end, and 16 inches at the other end? 

Ans. 84 feet. 
12 inches -f- 16 inches = 28 inches ; 28 inches —- 2 = 
mean width 14 inches. By the table 14 inches is 2^ times 
the length ; therefore 36 feet X 2^ = 84 feet. 

3. What is the value of a plank 24 feet long and 27 
inches wide @ 3 J cents per foot ? Ans. $3.92. 

4. Required the contents of a plank 18 feet long and 4 
inches wide ? Ans. ^ X § = ¥ = 12 feet - 

5. What are the contents of 1,860 feet running lengths 
of 2 inches X 2 inches ? Ans. 620 feet. 

Solution. — 1,860 -~ J= 620 feet. 

6. In 2,500 feet running lengths how many feet contents 
of 2 inches X 12 inches ? Ans. 5,000 feet or 5 M. 

2,500 feet X 2 = 5,000 feet, or 5 M. 



14 



SELF-INSTRUCTOR 



Table for Three-inch Deals, from 3 to 24 inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


3X3 = | 


3X 9==2i 


3X15 = 3J 


3 X 20 = 5 


3X4=1 


3 X 10 = 2j 


3 X 16 = 4 


3X21 = 51 


3X5=lJ 
3X6=lJ 


3X H = 2j 


3X 17 = 4j 


3 X 22 = 5j 


3 X 12 = 3 


3 X 18 = 4j 


3X23 = 5| 


3X7 = 1| 


3X 13 = 3j 


3X 19 = 4j 


3 X 24 = 6 


3X 8 = 2 


3 X 14 = 3j 







EXERCISE. 

1. What are the contents of a deal 3 inches thick, 6 
inches wide, and 30 feet long ? Ans. 45 feet. 

By the table 3X6 is 1^- times the length, for the con- 
tents ; therefore 30 feet X l£ == 45 = contents. 

2. What are the contents of a deal 3 inches X 12 inches 
and 331 feet long? Ans. 100 feet. 

3. In 2,700 feet of running lengths of 3 inches X 20 
inches, how many feet ? Ans. 13,500 feet. 

By the table 3 X 20 is 5 times the length, for the con- 
tents ; 2,700 X 5 = 13,500 feet. 

4. Required the number of feet running lengths of 3 X 4 
that will be equal to 2,000 feet running lengths of 3 inches 
X 10 inches ? Ans. 5,000 feet. 

5. What number of feet of running lengths of 2 X 3 
will be equivalent to 24,000 feet running lengths of 3 X 12 
inches. Ans. 144,000 feet. 

Solution. — By the table 3 X 12 is 3 times the length, for 
the contents ; therefore 24,000 feet X 3 = 72,000 feet = 
contents of 3 X 12 inches, and by the table 2 X 3 is = to 
half the length, for the contents ; therefore 2 X 3 is 2 times 
the contents for the running lengths, consequently 72,000 
feet X 2= 144,000 feet running length. 



ON LUMBER SURVEYING. 15 

Table for Four-inch Deals, from 4 to 12 Inches wide. 



Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


4X4 = 1| 


4X7 = 2^ 


4X 9 = 3 


4XH = 3f 


4X5=lf 


4X8 = 2f 


4X10 = 31 


4 X 12 = 4 


4X6 = 2 









EXERCISE. 

1. What are the contents of a deal 4X4 inches, and 20 
feet long? Ans. 26f feet. 

2. What are the contents of a deal 4x5 and 24 feet 
long? 

3. Required the contents of a deal 



long? 



Ans. 40 feet. 

4X6 and 26 feet 

Ans. 52 feet. 

4. Required the contents of a deal 4 inches X^ inches 
and SO feet long? Ans. 120 feet. 

5. What is the value of a piece of oak 36 feet long, 4 
inches thick, and 11 inches wide, @ 4^- cents per square 
foot? 

6. In 2,800 feet of running lengths of 4 inches X 12 
inches, how many feet of superficial measurement are there ? 

Ans. 11,200 feet. 

7. How many feet running lengths of 4 inches X 12 
inches deals are equal to 3,000 feet running lengths of 2 X 
6 ? • Ans. 750 feet. 

8. What is the amount of lumber in the following cargo, 
and its value @ $15.00 per M ? 

Surveyed from Bennett & Co., of Boston, Mass., to Ship 
Aurora, Capt. Jones, — 

2,758 pieces 2 X 8 and 16 feet long. 
3,800 pieces 4 X 12 and 30 feet long. 
2,600 nieces 4 X 10 and 16 feet long. 
250 M of Mer. spruce laths @ $2.50 per M. 

Ans. 653,497 feet of lumber. 250 M laths. 
Value of lumber, $9,802,451 
Value of laths, 625.00 



$10,427.45i 



16 



SELF-INSTRUCTOR 



Table of Five-inch Timber, from 5 to 12 Inches wide. 



Inches. Feet. 

5X6 = 2^ 
5X7 = 2H 
5 X 8 = 3 J 



Inches. Feet. 
5X 9 = 3f 
5X10 = 4 
5XH = 4j^ 
5 X 12 = 5 



Table of Six-inch Timber, from 6 to 12 Inches wide. 



Inches. Feet. 


Inches. Feet. 


6X 6 = 3 


6 X 10 = 5 


6X7 = 3j 


6 x H = 5j 


6X8 = 4 


6 X 12 = 6 


6X9 = 4 





EXERCISE. 

1. What are the contents of a piece of timber 5 inches X 
5 inches and 24 feet long ? Ans. 50 feet. 

By the table 5 X 5 is 2^ times the length, for the con- 
tents ; therefore 24 feet X 2 T V = ¥ X ff =W = 50 feet 
in board measure. 

2. Required the contents of a joist 5x8 and 30 feet long? 
30 feet X H = 100 feet. Ans. 100 feet. 

3. Find the contents of a beam 6 inches X 8 inches and 
36 feet in length ? Ans. 144 feet. 

36 feet X 4 =144 feet. 

4. How many running feet of 6-inch X 8-inch timber are 
equal to 3,500 feet running lengths of 5 X 1 2 inches ? 

Ans. 4,375 feet. 
By the table 5 X 12 is 5 times the length, for the con- 
tents, and 6X8 = 4 times the length ; therefore 3,500 feet 
X 5= 17,500 feet = contents of 5 X 12 ; then 17,500 -r- 
4 =s 4,375 feet = the number of feet in length of 6 X 
8 = 3,500 feet of 5 X 12. 



ON LUMBER SURVEYING. 



IT 



5. What will a beam cost 48 feet long, 6 inches by 11 

inches, @ 3J cents per foot ? Ans. $9.24. 

48 X 5 J feet = 264 feet = contents ; 264 X H cents = 

$9.24. 



Table of Timber from 7 X 7 to 12 X 20. 



Seven-inch Timber. 


Eight-inch Timber. 


Nine-inch Timber. 


Inches. Feet. 


Inches. Feet. 


Inches. Feet. 


7X 7=4^ 
7X 8 = 4§ 


8 X 8 = 5^ 


9X 9 = 6} 


8 X 9 = 6 


9X 10 = 7j 


7X 9 = 5^ 


8 X 10 = 6§ 
8 X H=7^ 


9XH = 8^ 


7 X 10 = 5f 


9 X 12 = 9 


7X11 = 6^ 


8 X 12 = 8 




7 X 12 = 7 






Ten-inch Timber. 


Eleven-inch Timber. 


Twelve-inch Timber. 


Inches. • Feet. 


Inches. Feet. 


Inches. Feet. 


10 X 10= 8± 

10x11= 9<r 


iiXn = io T 1 2 


12 X 12 = 12 


11 X 12 = 11 


12 X 14 = 14 


10 X 12 = 10 




12 X 16=16 

12 X 18 = 18 
12 X 20 = 20 



1. What are the contents of a piece of timber 12 by 12 
inches and 30 feet long? Ans. 360 feet. 

2. What are the contents of a beam 7 inches by 9 inches 
and 30 feet long ? Ans. 157J feet. 

3. Required the contents of a piece of timber 9 X 10 
inches and 40 feet long ? Ans. 300 feet. 

By the table 9 X 10 = 7J times the length ; 40 feet X 7J 
= 300 feet. 

4. In 2,500 feet contents of 9 X 10, how many feet run- 
ning lengths of 9 X 10, and of 11 by 12 ? 

Ans. Of 11 X 12, 227 T 3 T feet. Of 9 X 10, 333J feet. 

5. What is the cost of 2,000 feet running lengths of 12- 

2 



18 SELF-INSTRUCTOR 

inch by 20-inch timber @ 3 cents per foot of board measure ? 
Ans. $1,200.00. 

6. Required the contents of a piece of pine timber 8 inches 
by 12 inches and 24 feet long? Ans. 192 feet. 

7. What is the difference in feet of board measure be- 
tween 2,000 feet running lengths of 9 X 12 and 2,000 feet 
running lengths of 12 X 12 ? 

Ans. 12 X 12 is 6,000 feet more. 
By the table 12 X 12 = 12 times the length, and 9 X 
12 = 9 times ; therefore 12 — 9 = 3 feet difference ; 2,000 
X 3 = 6,000 feet difference. 

Example showing the Manner of Drawing or Ruling a 
Shingle for Plank or 2-inch, also the Mode of Dotting. 

Rule. — Take a shingle and rule it, as shingle No. 1 is 
ruled, the dimensions along the top column, and the lengths 
down the side column ; then take a pencil and make a dot, 
thus (,), for every plank, or deal, or piece of timber, as the 
case may be. Suppose I want to dot a 2 X 6, 22 feet long, 
3 times, I run along the top column of dimensions till I 
come to 2 X 6 ; I then go down said line till I come oppo- 
site 22 in the column of lengths, I then make three dots, 
thus (,.,). Then when I have finished dotting, I count all 
the dots, and place the figures as in the above shingle ; those 
figures I afterwards transfer to my specification, in order to 
find the contents of the whole quantity of pieces I have 
dotted. 

P. S. — You can, if required, rule your shingle so as to 
include any length or dimension, and most shingles are 
drawn as shingle No 1 is. 



ON LUMBER SURVEYING. 
Plank Shingle, No. 1. 



19 



Lengths, g g 


X 

18 


X 

4 

5 
4 
5 
3 

12 

4 

3 
2 

4 


X 

10 
5 

4 

1 

10 

6 

5 

1 
4 


CO 

X 

15 
5 

5 
2 

2 

7 
3 


X 

9 

7 

2 

2 

• 25 
1 

4 
4 


CO 

X 

2 
6 

5 

8 
5 

4 

2 
2 

3 


OS 

X 

12 

8 

6 
2 

3 

4 

1 
1 

3 


o 

X 

6 

8 

5 

10 

2 
3 

2 
2 


i— i 

X 

5 
2 

9 

4 

4 

5 
1 

.. 16 
1 

2 


X 

4 

1 

2 

10 

5 

6 

4 
7 

6 


12 


13 


9 

2 

5 

3 
15 


14 


15 


16 


17 


18 


19 


20 


21 


10 
9 


22 



20 



SELF-INSTRUCTOR 



Example of Specification of the Plank Shingle No. 1, 
showing the manner of finding the Contents. 

Rule. — One sixth of the length of 2-inch stuff multi- 
plied by the width will give the contents in feet of board 
measure or superficial feet. 



Specification 


0/ 


Plank 


Shingle 


JVb. 1. 


a 
era 

Dimen- 
sions. 


CO 

X 

<N 

18 

9 


<* 

X 

4 
5 


X 

10 
5 


CO 

X 

15 

5 


X 

9 

7 


GO 

X 

2 

6 


05 

X 

CI 

12 

8 


© 

X 

6 

8 


X 


1— 1 

X 


Contents. 


12 


5 

2 


4 
1 


1,120 


13 


834 


14 




4 


4 
1 

10 
6 

5 

1 


5 
2 

2 
7 


2 

2 

25 
1 
4 
4 


5 

8 
5 

4 
2 
2 

3 


6 




9 


2 


623 


15 


2 


5 
3 

12 


2 


5 
4 


4 




422 


16 


5 


4 


10 


1,000 


17 


lio 

i 

3 2 

4 3 


5 
1 

16 

1 
2 


5 

6 
4 

7 
6 


482 


18 


1,155 


19 


3 

15 

10 


4 
3 

2 


792 


20 


1 
1 

3 


2 
2 


1,180 


21 


885 


22 


9 


4 


4 


3 


828 




Total, 9,321 feet. 



Pule for calculating a 2-Inch or Plank Specification. 

Multiply the number of pieces or dots in each square of 
the table by the width of said pieces, and the product by J 
of the length for the contents. 



ON LUMBER SURVEYING. 21 

To find the Contents of Specification Shingle, No. 1. 

Multiply the number of pieces in each square of the table, 
opposite the first length, 12 feet, by the widths of the differ- 
ent numbers of said pieces, and then by ^ of the length for 
the contents ; thus, for the first column running parallel to 
the top of the shingle, 



eadth. 


3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


i. Pieces. 


18 


4 


10 


15 


9 


2 


12 


6 


5 


4 



54 16 50 90 63 16 108 60 55 48 

Then add all the products, 54+16 + 50 + 90 + 63 + 
16 + 108 + 60 + 55 + 48 == 560. Then 12, the length, 
■+ 6 = 2 feet, 560 X 2 = 1,120 = contents of the first 
column. Thus proceed until the contents of all the columns 
are found, then add the whole together for the total contents 
of the shingle. 

P. S. — In this treatise, when there is a fraction of half a 
foot over, it is called a foot ; when less than half a foot, 
nothing. 

For Joist or Scantling. 

Take the running lengths of the different dimensions and 
mark down every 100 feet, then add up your shingle, and 
multiply the different sums by the multiplier of each dimen- 
sion, as found in the tables for the contents of each. Hem- 
lock joist is generally, computed by this plan. 



22 



SELF-INSTRUCTOR 
Joist Shingle. 



2X3 


2jX3 


2X4 


2^X4 


2^X3 


2JX4 


3X4 

250 


100 


100 


10 


100 


100 


100 


100 


100 


90 


100 


100 


100 


250 


100 


100 


100 


100 


100 


100 


100 


50 


100 


100 


100 


100 


100 


100 


200 


100 


100 


100 


100 


100 


100 


100 


150 


100 


100 


100 




100 


100 


100 


100 


100 


100 






25 


100 


100 


100 


100 






150 


100 


100 


100 








200 


50 


100 


100 








100 




100 










613 


625 


667 


833 


450 


375 


900* 





* The numbers at the foot of the columns are feet of hoard measure. 



3 inches by 4 inches by the table is once the length, there- 
fore there are 900 feet of 3 X 4 contents. There are in 
the joist shingle 500 feet running length of 2^ X 4, and 
2 J X 4 is = | times the length ; therefore, 500 -f- f ±= 
to 375 feet = contents of 2£ X 4. There are 800 feet run- 
ning lengths of 2 \ X 3, and 2J X 3 is -^ times the length ; 
therefore, 800 — tfc = 450 = contents. There are 1,000 
feet of 2J X 4 ; therefore, as 2 J X 4 is f of the length, the 
contents will be equal to 1,000 -~ g = 833 feet. Of 2 
inches X 4 inches, 1,000 feet, which divided by f, will be 
the contents = 667 feet. Of 2^ X 3 there are 1,000 feet, 
and 2J X 3 is = f times the length ; therefore, 1,000 -f- § 
= 625 feet. Of 2 X 3 there are 1,225 feet running lengths, 
and 2 X 3 is ^ the length ; therefore, 1,225 
feet. 



i = 612J 



ON LUMBER SURVEYING. 
New York Deal Shingle, 3-Inch, No. 2. 



23 



Dimen- 
sions. 


CO 

X 

eo 


X 

eO 


00 

X 

09 . 


OS 

X 

eo 


O 

X 

eo 


X 

eo 


X 

eo 


14 


24 


14 




16 












20 


18 


20 


30 


15 


36 


20 


30 


11 


5 


3 

1 


4 


16 


12 




6 


8 




7 


10 


17 




9 




15 


15 


16 




24 
5 
3 


27 


18 


" 26 


8 




S 




9 
2 


10 


19 


21 


12 


1 


. 


4 


20 


12 


12 


4 


3 


4 


4 




10 


21 




16 




4 
5 


4 


4 
3 


5 




27 
3 


22 


8 


4 


4 


12 


23 






24 


3 


7 










27 


20 


20 


24 




15 


4 


4 

4 


8 


4 


4 


25 


6 
5 




4 


3 




8 


10 


9 


26 










4 
5 


6 


10 


10 


9 




20 


27 


10 




4 


5 


4 


4 


9 


28 






4 


5 


4 


4 


5 




18 


29 








5 








11 


9 


12 


10 


30 


30 


24 




8 


5 


4 


21 


9 


20 



P. S. — New York deal is from 12 feet up in length, and from 6 to 12 inches wide, 
be good spruce lumber, free from cracks, rots, or large knots, etc. 



24 SELF-INSTRUCTOR 

Specification of New York Deal Shingle, No. 2. 





CO 


»> 


00 


a 


o 


i— i 


<N 




Lengths. £ § 

g-a 


X 

CO 

24 


X 

CO 

14 


X 

CO 

20 


X 

CO 

16 


X 

CO 

18 


X 

CO 

20 


X 

CO 

30 


Contents. 


14 


4,571 


15 


36 


20 


30 


11 


5 


3 


4 


3,098 


16 


12 




6 


8 


10 


1 


7 


1,548 


17 




9 


24 


15 


15 


16 


27 


4,420 


18 


26 


8 


5 


9 


8 


7 


10 


2,744 


19 


21 


12 


3 


2 


1 


7 


4 


1,838 


20 


12 


12 


4 


3 


4 


4 


10 


2,095 


21 




16 


27 


4 


4 


4 


5 


2,667 


22 


8 


4 


3 


5 


4 


3 


12 


1,991 


23 


27 




20 


24 


3 


7 


20 


5,089 


24 




15 


4 


4 


8 


4 


4 


2,070 


25 


6 


10 


4 


4 


3 


9 


8 


2,494 


26 


5 


10 


10 


9 


20 


4 


6 


3,750 


27 


10 


9 


4 


5 


4 


5 


4 


2,315 


28 




18 


4 


5 


4 


4 


5 


2,429 


29 


30 


11 


9 


12 


5 




10 


4,401 


30 


24 


I 20 


8 


5 


4 


21 


9 


5,790 



JRule for finding the Contents of 3-Inch Deals. 

Multiply J of the length of the deals by the breadth of 
them, for the contents. 

This shingle is done the same way as the plank shingle 
No. 1, excepting that £ of the lengths are taken instead of £ 
of them. 



ON LUMBER SURVEYING. 
New York Deal Shingle, A-lnch, No. 8. 



25 



1 

Dimen- 
sions. 


X 


X 


QO 

X 


X 


o 

X 


X 


X 


14 




10 


••• 30 


12 


14 


17 


5 


32 


15 


24 


10 


11 




10 

5 


16 

4 


21 


16 


4 


7 


4 


5 


13 


17 






7 


•• 11 


14 


24 


18 


9 


18 


7 


2 




4 


3 


5 


4 


19 


7 


4 


4 


4 






5 


10 


20 


8 


6 




3 


2 


• 10 


12 


21 


• 25 




8 
7 


4 

7 


6 


5 
12 


4 


22 




6 


9 


11 


23 


8 


3 


2 


3 


1 


2 


3 


24 


6 


4 


5 


3 


3 


•• 11 


6 



Rule for finding the Contents of A-lnch Deals. 

Multiply the length divided by 3 by the breadth for the 
contents in feet of board measure. 



26 



SELF-INSTRUCTOR 



What are the contents of 32 pieces 14 feet long and 4X6? 
32 X 14 = 448 feet of running length, then 14 -h 3 = 4| 
= i of length of each piece. And 4X6 inches by the 
table is = 2 times the length, for the contents, therefore 
448 X 2 = 896 feet = contents. By taking J of the 
length, it is done thus, 32 pieces X 6, their breadth = 
192 X 4§ = 896 feet, contents. Or multiply the number 
of pieces by the length of one, and the product by ^ of the 
width of the deals for the contents of 4-inch. 



Specification of New York Deal Shingle, 4-Inch, No. 3. 



O rr 


CO 


j> 


00 


Ci 


© 


s 


C<J 




Lengths, g § 


X 


X 


X 


X 


X 


X 


X 


Contents. 


A ■ 


32 


10 


30 


12 


14 


17 


5 




14 


4,653 


15 


24 


10 


11 




10 


16 


21 


4,150 


16 


4 


7 


4 


5 


5 


4 


13 


2,133 


17 


9 




7 


11 


18 


14 


24 


4,709 


18 


7 


2 


1 


4 


3 


5 


4 


1,398 


19 


7 


4 


4 


4 


10 




5 


1,887 


20 


8 


6 




3 


2 


10 


12 


2,607 


2.1 


25 




8 


4 


6 


5 


4 


2,891 


22 




6 


7 


7 


9 


12 


11 


3,777 


23 


8 


3 


2 


3 


1 


2 


3 


1,380 


24 


6 


4 


5 


3 


3 


11 


6 


2,832 




Total, 32,417 feet. 





ON LUMBEK SUKVEYING. 



27 



Solution of Specification No. 3. 



No. Br. Products. 


No. Br. Products. 


No. Br. Products. 


32 X 6 = 


192 


9 X 6 = 


54 


8X6 = 


48 


10 X 7 = 


70 


8 X 7 = 


56 


6 X 7 = 


42 


5 X 12 = 


60 


11 X 9 = 


99 


3X9 = 


27 


30 X 8 = 


240 


18 X 10 = 


180 


2 X 10 = 


20 


12 X 9 — 


108 


14 X 11 = 


154 


10 X 11 = 


110 


14 X 10 = 


140 


24 X 12 = 


288 


12 X 12 = 


144 


17 X 11 = 


187 




















831 
5§ 




391 
6| 




997 


17 -r- 3 = 


20 -J- 3 = 


14 -r- 3 = 


4 


















Contents 


4,709 


Contents 


2,607 


Contents 


4,653 








7X6 = 


42 


25 X 6 = 


150 






24 X 6 = 


144 


2 X 7 = 


14 


8 X 8 = 


64 


10 X 7 = 


70 


1 X 8 = 


8 


4X9 = 


36 


11 X 8 = 


88 


4X9 = 


36 


6 X 10 = 


60 


io X io = 


100 


3 X 10 = 


30 


5 X H = 


55 


16 X 11 = 


176 


5 X H = 


55 


4 X 12 = 


48 


21 X 12 = 


252 


4 X 12 = 


48 








830 


■ 


233 


21 -T- 3 = 


413 

7 


15 -4- 3 = 


5 


18 — 3 = 


6 






Contents 


2,891 


Contents 


4,150 


Contents 


1,398 








6 X 7 = 
7X8 = 


42 
56 


4X6 = 


24 


7 X 6 = 


42 


7X7 = 


49 


4X7 = 


28 


7 X 9 = 


63 


4X8 = 


32 


4X8 = 


32 


9 X 10 = 


90 


5 X 9 = 


45 


4X9 = 


36 


12 X H = 


132 


5 X 10 = 


50 


io X io = 


100 


11 X 12 = 


132 


4 X H = 


44 


5 X 12 = 


60 








12 X 13 = 


156 








515 

'3 




298 


22 — 3 = 








400 

5^ 


19 -r- 3 = 


6* 






16 -f- 3 — : 


Contents, 


3,777 




r^Antpn tQ 


1,887 




Contents, 


2,1.33 


V/VM-ltt/ll tOy 







28 



SELF-INSTRUCTOR 



Solution 


of Specification No. 3. — 


- ( Continued.) 


No. 

8 X 
3 X 

2 X 

3 X 

1 X 

2 X 
3X 


Br. P 

8 = 

9 = 

10 = 

11 = 

12 = 

3 = 


roducts. 
48 
21 
16 
27 
10 
22 
36 


No. Br. 1 
6X6 = 
4X7 = 
5X8 = 
3X9 = 
3 X 10 = 

11 X 11 = 
6 X 12 = 

24 — 3 = 
Contents, 


'roducts. 
36 
28 
40 
27 
30 
121 
72 


24 feet being the 
length of the pieces 
in the last column, 
I take the J of it = 
8, and multiply it 
by the product of 


23 -5- 


180 

7 2 
'3 


354 

8 


the No. of pieces 
and their breadths. 


Contents, i^so 


2,832 





Rule for computing 5-inck Timber. 

Multiply the number of pieces in each square of the 
shingle, by their width as given in the top column, and the 
product by the length divided by 2f for the contents. 

By multiplying the length of a 5-inch stick by the width 
of the same, and the product by the length divided by 2f, 
you will get the contents in feet of Board Measure. 

Required the contents of 33 pieces 10 feet long of 5 
X5. 

1st Solution. — 33 X 10 = 330 X 2 T V= 6871 f ee t. 

2d Solution. — Find the contents of 10 pieces 33 feet 
long and 5 by 5.. 10 X 5 =50, 33 -~ 2f = T % X ¥ = 
\% =13|, therefore 50 X 13j= 687^ = Ans. 



ON LUMBER SURVEYING. 



29 



Timber 


Shingle Five-inch, No. 4 








00 

Dimen- 
sions. 


ift 

X 

1ft 
7 

8 

6 


X 

1ft 

2 
6 
5 


X 

l» 

5 
5 


00 

X 

1ft 
16 

8 


X 

»ft 


o 

X 

ift 


X 

1ft 
5 

4 

5 

2 

15 

4 

2 

3 


!-H 

X 

1ft 


20 


2 

5 
3 
5 


1 

5 
5 

8 


5 


21 


4 


22 


5 
2 
3 

1 
2 
6 


2 

5 
8 
2 
1 
3 


5 


23 


•••• 20 

2 

7 

7 


6 
7 
1 
3 
.... 12 

5 


8 


24 


3 

4 
4 
3 
3 
4 
3 


3 

4 
•• 10 
3 
3 
3 


9 


25 


5 


26 


1 


30 


2 


31 


3 
3 
2 


3 
2 
1 


6 
2 
5 


9 


32 


•• 10 


7 
4 


1 


33 


9 



30 



SELF-INSTRUCTOR 



Specification of Five 


-inch 


Timber Shingle, No. 4. 


, 


lO 


«5 


i> 


00 


cs 


o 








Lengths. | § 


X 


X 


X 


X 


X 


X 


X 


X 


Contents. 


5* 


7 


2 


5 


16 


2 


1 


LO 

5 


5 




20 


2,941 


21 


8 


6 5 


8 


5 


5 


4 


4 


3,167 


22 


6 


5 


5 


2 


3 


5 


5 


5 


2,778 


23 




6 


2 


5 


5 


8 


2 


8 


3,191 


24 


20 


7 


3 


8 


3 


3 


15 


9 


5,570 


25 


2 


1 


1 


2 


4 




4 


5 


1,865 


26 


7 


3 


2 


1 


4 


4 


2 


1 


2,004 


30 




12 


6 


3 


3 


10 


3 


2 


4,025 


31 


7 


5 


3 


3 


3 


3 


6 


9 


4,405 


32 




7 


3 


2 


4 


3 


2 


1 


2,387 


33 


10 


4 2 

1 


1 


3 


3 


5 


9 


4,345 








Total, 36,678 feet. 



Example, showing how to compute a 5-inch Specification. 



No. Br. 


No. Br. 


No. Br. 


No. Br. 


7 X 5= 35 


8 X 5 = 40 


6 X 5 = 30 


6 X 6 = 36 


2 X 6= 12 


6X 6=36 


5X 6= 30 


2 X 7=14 


5 X 7= 35 


5X 7 = 35 


5 X 7 = 35 


5 X 8 = 40 


8 X 16 = 128 


8 X 8 = 64 


2X 8 = 16 


5X 9 = 45 


2 X 9= 18 


5X 9 = 45 


3 X 9 = 27 


8 X 10 = 80 


1 x 10= 10 


5 X 10 = 50 


5 X 10 = 50 


2 X H = 22 


5 X 11= 55 


4 X 11=44 


5 X H = 55 


8 X 12 = 96 


5 X 12 = 60 


4 X 12=48 


5 X 12 = 60 




353 


362 


303 


333 


*20 — 2f 3= 8 J 


214-2|==8| 
Contents, 3,167 


22 -5- 2-g- = 9g- 


23-=-2f = 9 T V 


Contents, 2,941 


Contents, 2,778 


Contents, 3,191 



* 20 feet, the length of the pieces, divided by 2§, and the result, 8£, multiplied 
by 353 = 2,941 feet = contents of 20 ft pieces. 

Invert J£ = ft X ^ = Iff. = 8J 



ON LUMBER SURVEYING. 
Timber Shingle, Six-inch, No. 5. 



31 



ST 

D 
f 

Dimen- 
sions. 


CO 

X 

C© 


X 

CO 


QO 

X 

SO 


C5 

X 

CO 


o 

X 

CO 


i—i 

X 

CO 


X 

CO 


20 






3 

7 


10 


5 


10 


... 14 


18 


3 


21 


7 


4 


5 


9 


•••• 26 


22 


7 


6 


6 


5 


10 


11 


.... 15 


23 


7 


5 


3 


6 








11 


22 


.... 15 


24 




3 


4 


4 


3 


3 


5 


25 






6 


5 


4 


10 




20 


9 


26 


7 


3 


4 


5 


4 






11 


... 14 


27 


3 


3 


5 


5 


6 


10 


... 14 


28 


10 


3 


4 


4 


5 


6 


6 


29 


7 


6 


4 


5 


9 


3 


5 


30 


7 


4 


2 


6 


5 


5 





Rule for finding the Contents of 6-inch Timber. 
Multiply the number of pieces or dots by the width of 
said pieces, and then multiply the product by half the length 
of one of the pieces, for the contents. 



32 



SELF-INSTRUCTOR 



What are the contents of 18 pieces of 6 X 6, and 20 feet 
long? 18X6 = 108; 20 h- 2 = 10,108 X 10 = 1,080 feet. 
By the Table 6 X 6 is three times the length for the con- 
tents, therefore 20 X 18 = 360 feet running length, 360 feet 
X 3 feet = 1,080. Am. 1,080. 

So we find the same result by both rules. 

Specification of Timber Shingle, No. 5. 





O f-i <N 




1 . 


CO 


t^ 


GO 


o> 


H 


<~ ' 


rH 




Lengths. f § 


X 


X 


X 


X 


X 


X 


X 


Contents. 


gco 


CO 


CO 


co 


CO 


CO 


CO 


CO 




20 


18 


3 


3 


10 


5 


10 


14 


5,710 


21 




7 


7 


4 


5 


9 


26 


6,321 


22 


7 


6 


6 


5 


10 


11 


15 


6.358 


23 


7 


5 


3 


6 


11 


22 


15 


7,900 


24 




3 


4 


4 


3 


3 


5 


2,544 


25 


9 




6 


5 


4 


10 


20 


6,712 


26 


7 


3 


4 


5 


4 


11 


14 


6,097 


27 


3 


3 


5 


5 


6 


10 


14 


6,237 


28 


10 


3 


4 


4 


5 


6 


6 


4,718 


29 


7 


6 


4 


5 


9 


3 


5 


4,988 


30 


7 


4 


2 


6 


5 


5 


6 


4,755 
















Total, 62,340 feet. 



Examples showing how to compute the Specification No. 5 of 
6-inch Timber. 



Br. No. 


Br. No. 


Br. No. Br. No. 


6X 18=108 


7 X 7 = 49 


6 X 7= 42 7 X 6= 42 


7X 3= 21 


8 X 7 = 56 


7 X 6= 42 7 X 5= 35 


8 X 3 = 24 


9 X 4= 36 


8X 6= 48 8X 3= 24 


9 X 10 = 90 


10 X 5 = 50 


9 X 5= 45 9 X 6= 54 


10 X 5 = 50 


11 X 9= 99 


10 X 10=100 10 X 11 = 110 


11 x io = no 


12 X 26 = 312 


11 X 11 = 121 11 X 22 = 242 


12 X 14 = 168 




12 X 15=180 


12 X 15 = 180 


571 


602 


578 


687 


20-7-2= 10 


21-7-2= 10J 


22-7-2= 11 

Contents, 6,358 


23-7-2= llj 


Contents, 5,710 


Contents, 6,321 


Contents, 7,900 



ON LUMBER SURVEYING. 33 

What is the cost of a piece of pine timber 6 inches X 10 
inches, and 38 feet in length @ 3 J cts. per foot ? 

Ans. $6.65. 

Solution. — Length 38 -r- 2 = 19 ; 19 X by the breadth 
10 = 190 feet, contents. 190 feet @ 3J = $6.65. 

By the Second Rule. 6 inches X 10 inches = 5 times 
the length, for the contents, therefore 38X5 = 190 feet. 
190 feet XH cts. = $6.65. 

Rule for finding the Contents of 1-inch Timber. 

Multiply the width by the length, divided by If. 

Required the contents of a piece of timber 7X7 and 20 
feet long ? 

Divide the length, 20 feet, by If (20 -r- If = llf), and 
multiply the breadth, 7 inches, by the quotient, llf. 

1 If = 3^ ; 3^> x 7 — - 2|& = 81 f feet = contents in su- 
perficial feet. 

2d Operation. — By the table 7X7 is = to 4 T ^ times 
the length, for the contents, therefore 20 feet X 4 I J j = 81f 
feet = contents. 

Timber is often surveyed and the contents marked on 
each piece, and then put down on a shingle for contents in 
its proper column. 



34 



SELF-INSTRUCTOR 



Timber 


Shingle 


, Seven-inch. 


iVb. 6, erne? 


Specification, 


13 

Dimen- 
sions. 


X 


00 

X 

16 


C5 

X 


o 

r-H 

X 

5 

7 


X 
*>• 


i— t 

X 

1> 


Contents. 


20 


24 


7 


18 
8 
8 


16 


9,321 


21 


9 


6 


8 


6 


5,059 


22 


4 


4 
4 


4 


6 


8 


10,062 


23 


9 


7 


7 
6 


18 
9 


27 


10,062 
7,420 


24 


21 


7 


8 


8 


25 


4 


5 


6 


6 


4 


3,807 


26 


14 

5 


2 


6 


5 

7 


5 


1,493 


27 


6 


4 
4 


5 


7 
4 


5,197 


28 


8 
15 


4 


7 


8 


5,390 


29 


7 


„ 


4 


6 


2 


5,988 


30 


8 


4 


5 


8 


7 


8 


6,755 


31 


16 


12 


4 


2 


1 


3 


5,624 



ON LUMBER SURVEYING. 
Timber Shingle, Eight-inch. No. 7. 



35 



a,: 
Lengths. g = 

5* 


00 

X 

00 


X 

CD 


o 

X 

00 


pi 

X 

00 


X 

00 

7 


26 










12 
5 




12 


9 


18 


27 


5 


3 


... 
3 




8 

4 


28 


2 


3 


6 


5 


29 


5 


4 


3 


7 


5 


30 




3 


2 

2 




8 


10 


10 


31 




5 


2 


8 


32 


4 


1 


2 
5 


3 


4 

7 


33 


5 
3 


4 


7 


34 


2 


6 
3 


5 


2 


35 




5 






9 


36 




6 


7 




12 


9 


37 


7 








15 






32 


24 


21 



i?w/e for finding the Contents of 8 by 8 Timber. 
Divide the length by 1J, and multiply the quotient by the 
width of the timber for the contents in feet of boajd 



measure. 



36 



SELF-INSTRUCTOR 



Example showing how the first column of 8-inch specifi- 
cation is done. 



Br. No. pieces each 26 feet long. 

8X 12= 96 
9 X 18 = 162 

10 X 12 = 120 

11 X 9= 99 

12 X 7= 84 



561 



26-5-H= 14 



26 



11 3 

x 2 2* 



Invert the divisor, 
fX 2 T 6 =¥ = 14. 



7854 feet= contents. 
Specification Shingle, Eight-inch. No. 7. 



S <n 


CO 


OS 


o 


^ 


<N 




Lengths. | g 


X 


X 


X 


X 


X 


Contents. 


S a 


CO 


CO 


CO 


CO 


CO 




26 


12 


18 


12 


9 


7 


7,854 


27 


5 


5 


3 


3 


8 


4,392 


28 


2 


3 


6 


5 


, 4 


3,845 


29 


5 


4 


3 


7 


5 


4,698 


30 


10 


3 


2 


10 


8 


6,660 


31 




5 


2 


2 


8 


3,782 


32 


4 


1 


2 


3 


4 


3,029 


33 


5 


4 


5 


7 


7 


6,314 


34 


3 


2 


6 


5 


2 


4,103 


35 




5 


3 


9 




4,060 


36 


12 


6 




7 


9 


8,040 


37 


15 


7 


24 


21 


32 


38,406 












Total, 9,5183 feet. 



ON LUMBER SURVEYING. 
Timber Shingle, Nine-inch. No. 8. 



37 



a 
en 

CO 

Dimen- 
sions. 


OS 

X 


o 

X 

Ci 


3 
X 

Ci 


X 

Ci 

5 


26 


6 




6 


14 


27 






5 


5 
2 




12 


18 


28 


2 


3 


4 


29 


4 


3 


2 


4 


30 


6 


2 


5 


31 


8 


3 


4 


4 


32 


5 


4 


2 


15 


33 


2 


1 


5 


3 


34 


2 


2 

4 


4 


3 


35 


2 


5 


5 


36 

1 


6 


2 


1 


3 



Rale for finding the Contents of Nine-inch Timber. 

Divide the length by 1 J and multiply the quotient by the 
breadth of the stick for the contents. 

Required the contents of a piece of timber 9 X 12 inches 
and 26 feet long? 

26 ~ 1^=191 19|X 12 =234 = contents. 



38 



SELF-INSTRUCTOR 
Timber Shingle, Ten-inch. No. 9. 



Lengths. 


P oi 

£ a 

a. 2 

5" 


io x io 


10 X ii 


10 X 12 


26 






5 




13 


36 


27 


4 


5 


4 


28 




5 




11 


11 


29 


8 


3 


4 


30 


4 


6 


4 


31 


6 


2 


1 


32 




3 


3 






27 


33 


5 


5 


5 


34 


3 


2 


6 


35 


1 


2 


5 


36 






12 






25 


24 



ON LUMBER SURVEYING. 39 

Specification of Timber Shingle, Nine-inch. No. 8. 







o 


_ 


<N 




S3 -r. 






i - * 


>- • 




Lengths. | § 


X 


X 


X 


X 


Contents. 


5 " 


« 


OS 


OS 


C5 




26 


6 


14 


6 


5 


6,240 


27 


18 


12 


5 


5 


8,039 


28 


2 


3 


4 


2 


2,436 


29 


4 


3 


2 


4 


2,958 


30 




6 


2 


5 


3,195 


31 


8 


3 


4 


4 


4,510 


32 


5 


4 


2 


15 


6,888 


33 


2 


1 


5 


3 


2,945 


34 


2 


2 


4 


3 


3,009 


35 


2 


4 


5 


5 


4,541 


36 


6 


2 


1 


3 


3,267 












Contents, 48,028 feet. 



Specification of 


Timber 


Shingli 


?, Ten-inch. No. 9. 


a k 
Lengths. | § 


o 

X 


i— i 
X 


X 


Contents. 


a* 


o 


o 


o 
5 




26 


36 


13 


12,198 


27 


4 


5 


4 


3,297 


28 


11 


5 


11 


6,930 


29 


8 


3 


4 


3,891 


30 


4 


6 


4 


3,850 


31 


6 


2 


1 


2,428 


32 


27 


3 


3 


9,040 


33 


5 


5 


5 


4,587 


34 


3 


2 


6 


3,513 


35 


1 


2 


5 


2,683 


36 


12 


25 


24 


20,490 








Contents, 72,907 feet. 



40 



SELF-INSTRUCTOR 



Rule for Ten-inch Timber. 

Divide the length by 1£ and multiply the quotient by the 
breadth, for the contents in feet of board measure. 

Required the contents of a stick 36 feet long 10 inches by 
11 inches? 

36-^-1^ = 30, and 30 X 11 = 330 feet = contents. 

2d Solution. — By the table 10X H is 9£ times the 
length, for the contents ; therefore, 36 feet X $k — 330 feet 
= contents. 



Examples showing how 9 and 10 inch specifications are 
made out. 



Nine-inch. 

Br. Pieces. Pro. 

9 X 6= 54 

10 X 14 = 140 

11 X 6= 66 

12 X 5= 60 

320 
26~li= 191 

2880 
320 


Ten-inch. 

Br. Pieces. Pro. 

10 X 36 = 360 

11 X 13 = 143 

12 X 5= 60 

563 563 
2 21\ 


3)1126 


563 
1126 
375 


375 



12,198 feet 

Contents = 6240 Length, 26 -f- 1£ ; If = 

Length, 26 -h 1 J ; 1 i = f = inverted to g ; £ X ¥ 
i . Inverted = f ; j X ¥ = H~ = 21f . 
= ¥=19*. ' 

P. S. — All the specifications in this book are done in a 
manner similar to the specification of the Plank Shingle 
.No. 1. 



ON LUMBER SURVEYING. 



41 



Eleven-inch Shingle No. 10. 



Dimen- 
sions. 


X 

i—i 


X 


20 


24 




36 


21 


6 


4 


22 




' 3 


9 


23 


4 


2 


24 


5 


1 


25 


5 


3 


26 


1 


2 


27 


5 


3 


28 


5 


2 


29 


6 


4 


30 


5 


2 







d 








o 








© 




CV. 




© 




CO 




r s^ 




© 




-t-» 




r£3 




u 




o 




& 




.d 








(M 








rH 


5^ 


'S 




X 


i-C 


© 






§ 


rO 




r-l 


CM 


03 




'S 


*s 


+3 




§ 




^2 




d 




— - 




O 


*> 


© 






















^ 


O 
J3 




e» 


V 


cr 1 




o *; 


««9 

5a 






1!! 


s 

13 


.2* 
S 




:43 © 

° o 








§ ^ 




H3 
d 

c3 




•all 








<3 s^ 

«r-i rH 




J? 




-m i—i 
















© GO 

rd I— t 




© 


© 


8 II 




© 


d 


2* 




> 


•-* 






A 


X 


js + 






© 


O 



42 



SELF-INSTRUCTOR 
Timber Shingle, Twelve-inch, No. 11. 



Lengths. g § 


X 


X 


CO 

i—i 

X 


00 

X 

I— t 


o 

X 


20 






4 


4 




25 


16 


16 


21 


3 


2 


1 


1 


2 


22 


4' 


1 

1 


2 


2 


3 


23 


2 




2 


1 


24 


8 


3 


8 


1 


2 


25 


4 


1 


1 


3 


1 


26 


1 


2 


2 


1 


4 


27 


2 


3 


3 


2 


3 


28 


1 


2 


3 


4 


4 


29 


4 


3 


1 


1 


2 


30 


2 


2 


3 


2 


5 





i?w/e /or Twelve-inch Timber. 

Multiply the length by the width for the contents in feet. 
Required, the contents of 16 pieces of 12 X 20 inch timber, 
and 20 feet long ? 16 X 20 = 320. 320 X 20 = 6,400 
feet = contents in feet of board measure. 



ON LUMBER SURVEYING. 
Specification of Shingle No. 10. 



43 



i« 








Lengths, s § 


11 X 11 


11 X 12 


Contents. 


5 » 








20 


24 


36 


12,760 


21 


6 


4 


2,194 


22 


9 


3 


2.722 


23 


4 


2 


1,434 


24 


5 


1 


1,774 


25 


5 


3 


2,085 


26 


1 


2 


834 


27 


5 


3 


2,252 


28 


5 


2 


2,028 


29 


6 


4 


3,030 


30 


5 


2 


2,272 






Total, 33,285 





Specification of Shingle No. 


11. 






<M 


Tj" 


CO 

l-H 


00 


a 




Lengths. § § 


X " 


X 


X 


X 


X 


Contents. 


A " 


<N 


<N 


<N 




<N 




20 


25 


16 


4 


4 


16 


19,600 


21 


3 


2 


1 


1 


2 


2,898 


22 


4 


1 


2 


2 


3 


4,180 


23 


2 


1 




2 


1 


2,162 


24 


8 


3 


8 


1 


2 


7,776 


25 


4 


1 


1 


3 


1 


3.800 


26 


1 


2 


2 


1 


4 


4,420 


27 


2 


3 


3 


2 


3 


5.670 


28 


1 


2 


3 


4 


4 


6,720 


29 


4 


3 


1 


1 


2 


4,756 


30 


2 


2 


3 


2 


5 


7,080 












Total 


, 69,071 



44 



SELF-INSTRUCTOR 



Rule for finding the Contents, of Battens or Two-and-a- 
Half-inch Stuff. 



Inch. 


Inch. 




What are the contents 


2jX2= A 


2JX 8 = 


= lf 


of a batten 22 ft. long 2£ 


2jX3= | 


2jX 9 = 


= 1* 


inches by 12 inches ? 


2iX4= $ 


2| X 10 = 


= ** 


By this rule 2£ X 12 


2j X 5 = 1 A 


H x ii = 


= 2ft 


is = 2^ times the length, 


2 J X 6 = l£ 


2^X12 = 


= H 


for the contents, therefore 


2jX7 = l^ 






22 ft. X H = 55 ft. ^ws. 





Batten 


Shingle, No. 


12. 






era 

Dimen- 
sions. 


CO 

X 


X 


00 

X 


Ci 

X 

Hot 

4 

4 


o 

1— 1 

X 


X 

(M 

12 


X 


20 




15 


8 


2 


4 





.... 45 


21 


8 


3 


4 


3 


4 


3 


22 




4 


3 


3 


2 


1 
3 


3 


9 
4 


23 


3 


4 


4 
1 


3 


4 


24 


12 


8 


4 


2 


1 


4. 


12 


25 


6 


9 


4 


3 


3 


3 


2 


26 




8 


4 


3 


2 


1 


3 


.... 24 



ON LUMBER SURVEYING. 



45 



Rule for finding the Contents of Battens. 

Divide the length of the piece by 4f , and multiply the 
product by the breadth of the piece, for the contents in feet ; 
or multiply the length by the number given in the table for 
the contents. Ans. 30 feet. 

What are the contents of a batten 24 feet long 2J by 6 ? 

21 x 6, by the Table, is = to 1 J times the length ; 24 X 
11 = 30 feet. 

Second Solution. — 24 -f- 4f = 5 ; 5x6 = 30 feet. 

The specification is made out according to the last solution. 



Specification of Batten 


Shingle, 1 


Vo. 12. 




to 


t> 


00 


a 


© 

I— ( 


r-l 


<N 




Lengths, g § 


X 


X 


X 


X 


X 


X 


X 


Contents. 


5" 


45 


15 


8 


4 


2 


12 


4 




1 

20 


2,812 


21 


8 


3 


4 


4 


3 


4 


3 


1,080 


22 


9 


4 


3 


3 


2 


1 


3 


917 


23 


4 


3 


4 


4 


3 


3 


4 


1,073 


24 


12 


8 


4 


2 


1 


4 


12 


1,880 


25 


6 


9 


4 


3 


3 


3 


2 


1,276 


26 


24 


8 


4 


3 


2 


1 


3 


1,765 










Tot 


al, 10,803 



Random Shingle No. 13, for any Dimension. 

{Contents given in the Columns.) 



■S3 m 


■* 


00 


<M 


•<* 


os 


<M 


CO 


t- 


<M 


<M 


CO 


OS 


<M 


^15 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


Mpq 


<M 


cq 


<M 


cq 


CO 


CO 


O 


CD 


t- 


O 




<M 


N 




100 


700 


250 


150 


120 


120 


150 


250 


210 


210 


120 


120 


160 


200 


120 


200 


120 


210 


950 


210 


100 


120 


_ 


_ 


410 


250 


120 


120 


200 


210 


260 


160 


640 


800 


150 


100 


60 


160 


420 


210 


120 


600 


150 


250 


150 


300 


240 


120 


350 


320 


210 


20 


_ 


150 


312 


120 


150 


100 


100 


120 


275 


100 


940 


4W 


150 


120 


40 


500 


600 


200 


200 


120 


200 


100 


150 


868 


120 


ISO 


500 


210 


250 


36 


_ 


500 


100 


120 


210 


!50 


100 


100 


210 


200 


180 


120 


641 


100 


12 


250 


120 


100 


100 


100 


200 


100 


200 


450 


250 


200 


240 


120 


100 














150 


200 


120 


150 




150 


60 

























46 



SELF-INSTRUCTOR 



Method of keeping Shingle No. 13. 

The contents are found by the Board Rule and marked 
on each piece, and afterwards placed in the proper column in 
the shingle. 

What is the total number of feet of merchantable spruce 
lumber in Random Shingle, No. 13. Ans. 23,464 feet. 

Random Shingle, No. 14. 
{Running Lengths given in the Columns.) 



Q 


CO 


CO 


os 


W3 


«o 


CO 


t- 


OS 


S 


cm 


Contents of the 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


X 


whole. 


CN 

100 


80 


100 


200 


72 


120 


CO 

100 


120 


20 


CO 

120 


o 
50 




3X6= 1,045 


100 


60 


150 


100 


72 


100 


100 


100 


18 


150 


20 


2X10= 2,833 


25 


40 


210 


75 


60 


150 


120 


100 


16 


120 


40 


4X8= 4,186 


125 


20 


110 


60 


40 


100 


200 


40 


24 


100 


20 


4X9= 2,970 
5X5= 1,096 


100 


15 


200 


40 


18 


110 


110 


20 


20 


100 


100 


200 




_ 


20 


19 


70 


120 


100 


18 


200 


100 


5X6= 2,137 


100 


28 


150 


10 


20 


60 


150 


60 


16 


100 


100 


6X6= 3,525 
7X 7= 2,797 


100 


86 


120 


75 


70 


40 


100 


50 


19 


150 


- 


100 


72 


100 


100 


60 


20 


60 


40 


24 


250 


150 


7X 9= 1,118 


200 


150 


12i 


150 


40 


30 


40 


20 


20 


100 


200 


8X10 = 10,466 
10X12 = 10,500 


150 


96 


150 


60 


20 


20 


20 


15 


- 


120 


120 


400 


100 


160 


100 


15 


20 


25 


20 


18 


- 


- 




1700 


697 


1570 


990 


20 
526 


15 

855 


30 
1175 


685 


~213 


60 
1570 


150 
1050 






1£ 

X 3 


1* 


2§ 


3 


21 


>A 


3 


*A 


H 


6§ 


10 


Total . 42,673 ft. 


1700 


697 


3140 


2970 


1052 


1.710 


3525 


2740 


1065 


9420 


10500 




L133 


348 


1046 




44 


427 




57 


53 


1046 






2833 


1045 


4186 




1096 


2137 




2797 


1118 


10466 







ON LUMBER SURVEYING. 



47 





oooooooooooooooooooooooo 
oooooooooooooooooooooooo 
oooooooooooooooooooooooo 




No. of feet in 
Length = to 
1,000 feet of 
Contents. 


_Ji-i _>-< "sH it- 

lO|f r-<|<» ~rH *-|o»r-<(coWl|eO0C^aae^|O9 »h|M*H»-<*4« |i- , H««>l l = ! H6» l«- 

ooiooeooo©i^co<oooi— iOio»:cot)((nnionoo) 

oocoococotooNcoocoTjicijiNntOTtiiNWeoajH 

oaiooiciTji^TjiNcoocio 

i— 1 tH i— ( 


» a 
0.2 

P 00 


lOOMJOejOHNM^OCOOOO^OtOW^TlirtiOOO 

xxxxxxxxxxxxxxxxxxxxxxxx 

iMOHNMNOllNININiMCKNHlNlNHHrHlMw-r-MCO 




o 

.s o-i 
jiil 

111 


oooooooooooooooooooooooo 
oooooooooooooooooooooooo 
oooooooooooooooooooooooo 

iniiiiiiiiiiiiiiiiiiiiiiiiiiiiii 

_H 

»l*- _Ji-t Jt- J09 .(r-t Ij<NO|r-l 
CO|t-CDlt-cO]l~r-llO(Nl»3 ^|rH ^|e»-'|e3 r >s r H|3» HrH H^rH:-,H<»e3|t-,-<|e*0|ca O O 

t-(iO(Mt-?OOcr.iCCOCOi-HrHOC50asOCCi— iG^iOOOO 

NLTTfOOCOiOCOINiiOJINr! (NO OOOlOONttlOiOOIN 


is 

E.S 

5* 


Or- i<NOQC.©i-l<NCi©i~i<MOi-l<Mi-<(N<M-«tf<vOOOOCOTi< 

XXXXXXXXXXXXXXXXXXXXXXXX 

I>NM»XCOOOOOO)OOCiOOOhh(N(NIN(NN H«H« 
.-lr-li-li^lT-li-lT-li-li-ti-l<?4(N 




s 

§ 


OOOOOOOOOOOOOOOOOOOOOOOO 
OOOOOOOOOOOOOOOOOOOOOOOO 

oooooooooooooooooooooooo 


No. of feet in 
Length = to 
1,000 feet of 
Contents. 


-Jr-I ->-l Ui ^(|aj OH 

00i0C0O(NOOO(MOOOG0OC0iOO<MOi-H<X)->^Tt'O 

(NNcijo^oooo^ooiiHomxioNoaifflTiirift 

■^«MWN(N^^CCiCOlM(N<N<7J0O(N<M(NNHH(M(NrH 


u 

5 o 

5" 


l>00OOH(NiaON0CI05OH(M(8N00ftOHNN000) 

XXXXXXXXXXXXXXXXXXXXXXXX 

TJ*T(f^rf(Tt(TfiOiOiOiO»OOiOiOOO^COCOCOi»NNI> 




a 
-2 

a . 
o 


OOOOOOOOOOOOOOOOOOOOOOOO 
OOOOOOOOOOOOOOOOOOOOOOOO 
OOOOOOOOOOOOOOOOOOOOOOOO 


No. of feet in 
Length ss to 
1,000 feet of 
Contents. 


,-i]l-. «|M "V< th|09 w]MW|l-- "*)« *"VH» 
OOOOOt>-OC£>O»0O00O©OT-iO-t<OC0C0OOO 
OOOOOiO'OCOO^OeOOOOt^O-^OCDCOiOOO 

oo>o(Nooo^«)«oou:«ooo«>iaoTiiTi(cont>«iio 

05NHHH l-l rH 


C 7- 

2 a 
B.2 
5" 


(NCOtfiCWNOOeiOHNOJ^iOOSOOffiOHIN^iOCO 
i-l iH .-1 ,-4 tH i-l 

XXXXXXXXXXXXXXXXXXXXXXXX 

M(NN<N(M(N(N(NN(NiNincoo:commmmcocoTfrf<f 



48 



SELF-INSTRUCTOR 



Rule showing how Table B is calculated. 
Divide the area or contents of the end into the given 
number of feet of contents, and the quotient will be the 
number of feet of running lengths, equivalent to the given 
number of feet of contents. 

1. What number of feet in length of 10 inches by 12 
inches will be equal to 1,000 feet contents. 

By the table 10 inches X 12 inches is 10 times the length, 
for the contents ; therefore, 1,000-7-10=100 feet in length. 

2. How many feet of 2 X 3 are equal to 1,000 feet of 
contents ? 

2 X 3 = \ the length ; therefore, 1,000 X 2 = 2,000 
feet = length required. 

Table C. — Number of Feet of the following Dimensions of Timber 
that will make 1,000 Feet, Cubic or Solid Measurement, 



5 X 

5 X 

5 X 

5 X 

5 X 

5 X 



No. of Feet 


Cubic 


in Length. 


Feet. 


5,760 


1,000 


4,800 


1,000 


4,114f 


1,000 


3,600 


1,000 


3,200 


1,000 


2 880 


1,000 


2,618 T 2 T 


1,000 


2,400 


1,000 


4,000 


1,000 


3,428^ 


1,000 


3,000 


1,000 


2,666| 


1,000 


2,400 


1,000 


2,181 T 9 T 


1,000 


2,000 


1,000 


2,938ft 


1,000 


2,57lf 


1,000 


2,285f 


1,000 


2,057| 


1,000 


l,870j$ 


1,000 


— 


1,000 



Dimensions. 



7X 
8 X 
8 X 
8 X 
8 X 

8 X 

9 X 
9 X 
9 X 
9 X 

10 X 
10 X 
10 X 



11 X 

ii X 

12 X 
14 X 16 
16 X 18 
18 X 20 
20 X 22 
22 X 24 



No. of Feet 
in Length. 



l,714f 

2,250 

2,000 

1,800 

1,636^ 

1,500 

1,777$ 

1,600 

1,455 T \ 

1,333^ 

1,440 

1,309-^ 

1,200 

1,190 T L° T 

i,090^r 
1,000 

642f 

500 

400 

327 T \ 

272 T 8 T 



No. of ft, 
of Cubic 
Measure. 



1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,C00 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 
1,000 



ON LUMBER SURVEYING. 



49 



Rule showing how Table C is computed. 

Multiply the breadth and width in inches together, and 
divide the product by 144, the number of inches in a square 
foot, and the quotient divided into the given number of cubic 
feet will give the number of feet in length, equal to said 
number of feet. 

How many feet running length of 6 inches X 6 inches are 
equal to 1,000 cubic feet? Arts. 4,000 feet. 

6 X 6= 36 ; 36 -f- 144 = T \ 6 ¥ = £ ; J inverted = to f 
X i-Oj- - = 4 -° T °-° == 4,000 feet of running lengths = 1,000 
cubic feet. 



Table showing the Numbers to multiply the Lengths of the 
following Dimensions by in order to find the Contents 
in Cubic Feet. 



Dimension. No. 


Dimension. No. 


Dimension. 


N T o. 


5X 5 = t¥4 


7 X 


ii = T y ? 


12 X 16 = 


1* 


5 X 6 = A 


7 X 


12 = A 


13 X 14 = 


m 


5X7 = f& 


8 X 


8 = 1 


. 14 X 16 = 


if 


5 X 8 = A 


8 X 


•=* 


16 X 18 = 


2 


5 X 9 = A 


8 i X 10 = | 


16 X 20 = 


2f 


5 X 10 = f f 


8 X 


« = « 


18 X 20 = 


H 


5xn = m 


8 X 


12 = 1 


20 X 22 = 




5X12 = A 


9 X 


»= 'A 


22 X 24 = 


H 


6 X 6 = ± 


9X 


10 = 1 


24 X 26 = 


4 


6 X 7 = A 


9 X 


ii = H 


26 X 28 = 


5tV 


6 X 8 = i 


9 X 


12 = } 


28 X 30 = 


5| 


6 X 9 = f 


10 X 


io = U 


30 X 32 = 


«* 


6 X 10 = A 


io X 


n = « 


32 X 34 = 


H 


6Xll = H 


io X 


12 = f 


34 X 36 = 


8* 


6 X 12 = J 


11 X 


11 = Hi 


36 X 38 = 


9* 


7X 7 = ^ 


ii X 


12 = h 


38 X 40 = 


lOf 


7X 8 = ft 


12 X 


12 = 1 


40 X 42 = 


Hi 


7X 9 = A 


12 X 


14 = 1± 


42 X 44 = 


m 


7 X 10 = f f 











50 SELF-INSTRUCTOR 

QUESTIONS FOR EXERCISE. 

1. Required the number of solid feet in a timber 6 inches 
X 6 inches and 40 feet long? Ans. 10 feet. 

Solution. — 6 X 6 = i of length, therefore £ of 40 = 
10 feet. 

2. What is the solidity of a piece of 6-inch X 12-inch 
timber 72 feet long? Ans. 36 feet. 

By the table 6 X 12 = J the length, for the contents; 
therefore J X 72 = 36 feet. 

3. What number of cubic feet are there in a piece of 
timber 40 feet long, 22 inches X 24 inches ? 

Ans. 146f feet. 

4. Required the number of feet in a piece of timber 32 
feet long, 5 inches X 12 inches? Ans. 13 J feet. 

Solution. — 32 feet X t 5 2 — ^i ^ eet — contents. 

5. What number of cubic feet in the following pieces, 
namely, 6 pieces 60 feet long 12 inches X 16 inches, and 12 
pieces 35 feet long and 16 inches X 18 inches? 

Ans. 15,840 feet. 

6. What are the contents in cubic feet of 6 pieces of 20 
inches X 24 inches and 35 feet long ? 

Ans. 11 If cubic feet. 

7. What number of cubic feet in a piece of timber 28 
inches X SO inches and 60 feet long? Ans. 350 cubic feet. 

Solution. — 60 X 5f = 350 feet of cubic measure. 

8. Required the contents in cubic feet of a piece of pine 
timber 30 inches X 32 inches and 30 feet in length? 

Ans. 200 feet. 

9. How many tons of timber (allowing 42 cubic feet to 
the ton) in a piece of timber 38 inches X 40 inches and 45 
feet long? Ans. Uptons. 

10. What will be the cost of a piece of pine timber 18 
inches X 20 inches and 30 feet in length @ 30 cents per 
cubic foot? Ans. $22.50. 



ON LUMBEE SURVEYING. 



51 



Rule to reduce Feet of Board Measure to Cubic Feet. 

Divide the contents in superficial feet by 12, and it will 
give the number of cubic feet ; or multiply the number of 
cubic feet by 12 and the product will be feet of board 
measure. 

In 1,200 feet of board measure how many cubic feet are 
there ? Ans. 100 cubic feet. 

Solution. — 1,200 -~ 12 = 100 cubic feet. 

Eequired the number of feet of board measure in 100 
feet of cubic measure ? Ans. 1,200 feet. 

100 X 12 = 1,200 feet of board measure. 



Second Method of making out a Specification. 

3-INCH SPECIFICATION BY THE SECOND METHOD. 



§ i 


CO 


i> 


00 


C3 


o 


i— i 


n 


*- 


Lengths, g § 
5' S 


X 


X 


X 


X 


X 


X 


X 


Contents. 


CO 

2 


CO 

3 


CO 

4 


CO 

6 


CO 

8 


co 
4 


CO 

6 




14 




15 


4 


2 


1 


4 


2 


8 


4 




16 


2 


4 


2 


1 


3 


2 


4 




17 


6 




1 




1 


3 


2 




18 


8 


4 


6 


1 


3 


2 


4 




19 


2 


1 


2 


3 


2 


4 


6 




20 


3 


2 


1 


4 


2 


1 


3 




21 


6 


4 


8 


2 


1 


3 


2 




22 


1 


5 


4 


3 


2 


1 


1 




23 


2 


1 


10 


4 


1 


2 


1 




24 




6 




4 




3 


2 




25 


4 




2 


8 


6 




4 




26 




3 


2 


1 




2 


8 




27 


6 


5 


1 




3 




2 




28 




8 




2 




4 


6 




29 


3 
1510 


1864 


5 
2092 


2 
2140 


1 
1717 


6 
2563 


4 

3807 






15,693 feet. 



52 



SELF-INSTRUCTOR 



Second Rule for Specifications. 
Multiply the number of pieces or dots in each square of 
the specification by the length of one of the pieces ; and 
multiply the product thus found by £ of the breadth of said 
pieces for the contents in board measure of 3-inch deals ; by 
^ of the breadth for 4-inch ; by J of it for plank, etc. 

Example showing how to make out the Three-inch Specifi- 
cation by Second Method. 



First Column 6 inches wide. 


Second Column 7 inches wide. 


14X2= 28 


14X3= 42 


15X4= 60 


15 X 2 = 30 


16 X 2 = 32 


16X4= 64 


17 X 6 = 102 


18 X 4 = 72 


18 X 8 = 144 


19X1= 19 


19 X 2 = 38 


20X2= 40 


20 X 3 = 60 


21X4= 84 


21 X 6 = 126 


22 X 5 = 110 


22 X 1 = 22 


23 X 1 = 23 


23 X 2 = 46 


24 X 6 = 144 


25 X 4 = 100 


26 X 3 = 78 


27 X 6 = 162 


27 X 5 = 135 


29 X 3 = 87 


28 X 8 = 224 


1,007 


1,065 


»* 


If 


1,007 


1,065 


503 


799 


Contents, 1,510 feet. 


Contents, 1,864 feet. 


6 inches, the breadth, di- 


7 inches, the breadth, di- 


vided by 4 is = to 1 J, and 


vided by 4 is = to If, and 


H X 1,007 = 1,510, the 


If X 1,065 = 1,864 feet = 


contents. 


contents. 



English deal specifications are generally made out by the 
second method. Both rules will give the same results. 



ON LUMBER SURVEYING. 53 

Specification of Philadelphia Deal Shingle. 



Lengths, g § 



14 
16 
18 
20 
22 
24 
26 



X 


Contents. 


05 




40 
35 
30 


1,680 
1,680 
1,620 


11 


660 


9 


594 


21 
6 


1,512 

468 



28 



34 
36 

38 
40 



X 



14 
8 
4 
4 
7 
14 
14 



Contents. 



1,176 
720 
384 
408 
756 
1,596 
1,680 



Contents, 14,934 feet. 



Philadelphia Deal Shingle. 



ft 
a 

Dimen- 
sions. 


I-H 

X 

CO 


Dimen- 
sions. 


l-H 

X 

CO 


14 




28 






14 


40 


16 




30 


8 




35 


18 




32 


4 




30 


20 




34 


4 


11 


22 




36 


7 


9 


24 




38 






14 


21 


26 


6 


40 




14 



54 SELF-INSTRUCTOR 

The specification of Philadelphia deals is done the same 
as the 3-inch specification ; or multiply the running lengths 
by 3 for the contents in feet of board measure. Philadel- 
phia deal is generally 12 inches wide and even lengths, from 
14 feet up, and the best quality of spruce lumber. English 
deals generally comprise all deals too short, or not good 
enough for Philadelphia or New York deals. Also short 
timber, battens, and plank, not suitable for other markets, go 
into the English deal pile. Deals that are knotty, cracked 
by the sun, or stained, or having wanes on them, and not 
poor enough for refuse, go to the English deal pile. New 
York deal must be the best quality of spruce, from 14 feet 
long up. 

Directions showing how to measure all hinds of Lumber 
by the Board Rule. 
Lay your rule across the board to be measured, at right 
angles to the further edge of the board, and let the outside 
edge of the board and further end of the rule be both even 
on that side, then observe the length of your board and turn 
your rule to the same length, then look on the line or column 
of that length, and you will find the contents marked on the 
rule just over the inside edge of the board. 

EXAMPLES FOR PRACTICE. 

1. What are the contents of a 1^-inch board 16 feet long 
and 12 inches wide ? Ans. 20 feet. 

By the rule the contents given for 1-inch board is 16 feet 
contents, to which add \ of the contents, which will give 
the contents for 1^-inch boards. 16-^-4=4; 16-f-4 = 
20 feet contents. 

2. What are the contents of a board 32 feet long and 12 
inches wide ? Ans. 32 feet. 

As there is no 32 on my rule, I find the contents by the 
rule of a board, half the length to be 1 6 feet ; which being 
doubled, gives the contents required = 32 feet. 

3. What are the contents of a 1^-inch board 20 feet long 
and 12 inches wide? Ans. 30 feet. 



ON LUMBER SURVEYING. 55 

By the rule an inch board 20 feet long and 12 inches 
wide will contain 20 feet, to which add half of 20 for the 
contents of a l^-inch board. 20 -J- 2 = 10 ; 20 + 10 = 30 
feet. 

4. Required the contents of a plank 24 feet long 2 inches 
X 12 inches ? Am. 48 feet. 

By the board rule, in a board 24 feet long 12 inches wide 
and 1 inch thick there are 24 feet, and as plank is 2 inches 
thick, therefore twice the contents of the face of it will be 
equal to the true contents, 24 X 2 = 48 feet. 

Rule for any Dimension. 

Multiply the number of feet in the face of the piece to 
be measured, by the thickness in inches, and it will give the 
contents in feet of board measure. 

Rule for measuring Logs or Round Timber. 

Multiply the length, taken in feet, by the square of one 
fourth of the mean girth, taken in inches, and this product 
divided by 1 44 will give the contents in cubic feet. 

Note. — The girth of tapering timber is usually taken 
about one third the distance from the larger to the smaller 
end. The rule is that in common use, though very far from 
giving the actual number of cubic feet ; 40 cubic feet as 
given by the rule are in fact — 50 T 9 n 2 o true cubic feet. 

EXAMPLE. 

1. How many cubic feet in a stick of timber which is 40 
feet long, and whose girth is 60 inches ? Ans. 62^ feet. 

60-^4=15 inches = i of girth; 15 X 15 = 225 = 
square of quarter of the girth ; 225 X 40 feet = 9,000 ; 
9,000 -j- 144= 62^ cubic feet. 

2. How many cubic feet in a piece of timber 21 feet long, 
and whose girth is 36 inches ? 

3. What are the contents of a log 100 feet long, and 
whose girth is 150 inches? 



56 SELF-INSTRUCTOR 

To find the largest Square Piece of Timber that may be 
sawed from a Round Stick of Timber, having the Diam- 
eter or Circumference of the Small End given. 
Rule 1. — Multiply the given diameter by .707106, or, 

multiply the given circumference by .225079. Or, as the 

diameter of a circle is equal to the diagonal of the inscribed 

square — 

Ride 2. — Square the diameter 

and take half the sum of the square, 

and extract the square root of it, 

ai.d the root thus found will be the side of the inscribed 

square. 

EXAMPLE. 

1. I have a piece of timber 30 inches in diameter ; how 
large a square stick can be hewn from it. 

By the last rule 30 squared = 30 X 30 = 900 ; 900 -^ 2 
= 450 ; «/4§ ^ = 21.21 -|- inches square. 

2. How large a square stick may be hewn from a piece 
of round timber 120 inches in circumference:? 

3. How large a square stick may be sawn from a piece 
of round timber 60 inches in diameter ? 

Having the Side of a Square Stick given^to find the Diam- 
eter of the Tree from which it was sawn. 

Rule. — Square the side and double it, and out of the 
product extract the square root. 

What must be the diameter of a tree that when hewn 
shall be 18 inches square ? Ans. 25 .44 inches. 

Table. 
12 lines = 1 inch. 
12 inches = 1 foot. 
3 feet = 1 yard. 

Inches multiplied by inches produce 
Parts marked thus '. 
Parts by parts give fourths, marked thus ,nf . 



ON LUMBER SURVEYING. * 57 

Inches are marked f . 
144 square inches make 1 square foot. 
9 square feet = 1 square yard. 
1,728 cubic inches = 1 cubic foot. 
50 cubic feet = 1 load. 
40 cubic feet = 1 ton of timber. 
16 cubic feet = 1 cord foot. 
8 cord feet, or 128 cubic feet = 1 cord of wood. 
1,980 feet superficial = 1 St. Petersburg standard of 
deals. 

Form of a Bill of Lading of Timber, Shingle No. 8, etc., etc. 

Shipped, in good order and condition, by Edmond B. 
Sanderson & Co., on board the good ship " Southern," 
whereof James Brown is master for this present voyage, 
now lying in the port of New York, U. S., and bound for 
Liverpool, England. To say : — 

47,928 ft. Mer. spruce, all under deck, 
100 M spruce laths, all under deck, 
80 M ft. Mer. pine, all on deck, 

being marked and numbered as in the margin ; and are to 
be delivered, in like good order and condition, at the afore- 
said port of Liverpool (the danger of the seas and fire 
always excepted), unto David Belt & Sons, or to assigns, 
he or they paying freight for the said timber at the rate of 
ten dollars per M feet, and one dollar per M for laths, with- 
out primage and average accustomed. 

In witness whereof the master of the said vessel hath 
affirmed to three bills of lading, all of this tenor and date ; 
one of which being accomplished, the others to stand void. 

James Brown. 

Dated at New York, U. S., 
May the 3d, A. D. 1870. 



58 SELF-INSTRUCTOR 

Bill of Lading. 



Shipped, in good order and condition, by T. 
Pandol & Co., on board the good schooner called 
the " Northern Dawn," whereof Daniel E. Bloomer 
is master for this present voyage, now lying in the 
port of Bangor, Me., and bound for New York. 
To say : — 

110 M feet hemlock lumber, all under deck, 

75 M feet spruce lumber, all on deck, 
120 M laths, all on deck, 



being marked and numbered as in the margin ; and 
are to be delivered, in like good order and condition, 
at the aforesaid port of New York (the danger of 
the seas and fire only excepted), unto Messrs. Den- 
ton and Beeters, or to assigns, he or they paying 
freight for the said lumber at the rate of four dol- 
lars per M feet, and sixty cents per M for laths, 
without primage and average accustomed. 

In witness whereof, the master of the said vessel 
hath affirmed to three bills of lading, all of this 
tenor and date ; one of which being accomplished, 
the others to stand void. 

Daniel E. Bloomer. 

Dated at Bangor, Me., 
June the 3tf, 1869. 



Surveyor's Bill for Services rendered. 

Bangor, Me., June the 2d, 1869. 

Messrs. Dunton & Boomer, 

To Daniel E. Shaw, surveyor, Dr. 
For surveying 250 M ft. of spruce lumber to 

schooner "Juno," @ 25c. per M . $62.50 



ON LUMBER SURVEYING. 59 

Survey Bill of Lumber, etc. 

Surveyed from James E. Dale & Sons, of Clinton, Iowa, to 
schooner " Pallas," Captain Dunn. To say : — 

36,500 ft. 2X6, from 12 ft. long up (mch.), spruce. 
35,600 " No. 1 pine boards. 
22,400 " hemlock boards (mch.). 
15,000 " 8 X 10 Mer. pine timber. 
250 M No. 1 pine shingles. 

Thomas B. Proudfoot, 

Surveyor. 
Clinton, Iowa, 
June the 12th, Anno Domini 1869. 

Surveyor's Receipt. 

$62.50. Bangor, Me., June the ±th, A.D. 1869. 

Received from Messrs. Dunton & Boomer sixty-two 
dollars and fifty cents, which pays for surveying 250 M feet 
of spruce lumber to schooner " Juno," @ 25c. per M. 

Daniel E. Shaw, Surveyor. 

NOVEL RULES 

For finding the Contents of Plank, Deal, Battens, Joist, and 
Timber, by multiplying a Fractional Part of the Length 
by the Breadth. 

2-inch is £ of the length multiplied by the breadth, for the 
contents. 

3-inch is \ of the length multiplied by the breadth, for the 
contents. 

4-inch is £ of the length multiplied by the breadth, for the 
contents. 

5-inch is the length divided by 2f , and the quotient multi- 
plied by the breadth. 



60 SELF-INSTRUCTOR 

6-inch is J of the length multiplied by the breadth, for the 

contents. 
7-inch is the length divided by If, and the quotient multi- 
plied by the breadth. 
8-inch is the length divided by 1^, and the quotient multi- 
plied by the breadth. 
9-inch is the length divided by 1^, and the quotient multi- 
plied by the breadth. 
10-inch is the length divided by 1£, and the quotient multi- 
plied by the breadth. 
11-inch is the length divided by 1^, and the quotient multi- 
plied by the breadth. 
12-inch, multiply the length by the width, for the contents. 
2^-inch, or battens, is the length divided by 4£, and the 

quotient multiplied by the breadth. 
p. S. — The above rules give the contents in feet of board measure. 

EXAMPLES FOR PRACTICE.' 

1. Required the contents in superficial feet of a piece of 
timber 10 inches X 12 inches and 40 feet long. 

Ans. 400 feet. 
Solution. — By the table, 10 inches is 1^ of the length mul- 
tiplied by the breadth. Therefore 40 feet -f- 1£ = -\° X 
| = 20.0 — 33£ ; 33i x 12 = 400 feet. 

2. What are the contents of a piece of timber 12 inches 
X 20 inches, and 40 feet long ? Ans. 800 feet. 

Solution. — 40 X 20 = 800 feet. 

3. What are the contents of a plank 2 inches X H inches 
and 36 feet long? Ans. 66 feet. 

Solution. — 2 inches is £ of the length. Therefore 
36 -r- 6 = 6 ; 6 X 11 = 66 feet. 

4. What are the contents of a piece of timber 8 inches X 
11 inches and 40 feet in length? Ans. 293^ feet. 

Solution. — 40 -v- H = -Y-; *r°-X| = \°- = 26f ; 26| 
X 11 = 293£ feet. 



ON LUMBER SURVEYING. 61 

Given the Breadth of a Rectangular Plank in Inches, to 
find how much in Length will make a Foot, or any other 
required Quantity. 

Rule. — Divide 144. or the area to be cut off, by the 
breadth in inches, and the quotient will be the length in 
inches. 

1. If a board be 6 inches broad, what length of it will 
make a square foot ? Am. 2 feet. 

Solution. — 144 inches -f- 6 inches = 24 inches; 24 
inches -4-12 inches = 2 feet. 

2. If a plank be 2 inches X 8 inches in size, what length 
of it will make 4 square feet ? Am. 3 feet. 

Solution. — 2X8 = 16, area of the end ; 144 -f- 16 = 9 
inches for 1 foot, which, being multiplied by 4 = 4 X 9 = 
36 inches = 3 feet. 

To find the Solid Contents of a Piece of Timber tapering 
regularly. 

Rule. — Multiply the sum of the breadths of the two ends 
by the sum of the depths, to which add the product of the 
breadth and depth of each end ; \ of this sum, multiplied by 
the length, will give the exact solidity of any piece of 
squared timber tapering regularly. 

1. How many feet in a piece of mahogany whose ends 
are rectangles, the length and breadth of one being 14 and 
12 inches, and the corresponding dimensions of the other 
end 6 and 4 inches ; also the length 30-J feet ? 

Am. 182 2 t cubic feet. 
Solution. — 
14 + 6 = 20 12X14=168 

12 + 4 =16 6 X 4= 24 

20 X 16 = 320 

512 sq. in. = % 2 - sq. ft. 
Then \ X V X 30£ = l&fr cubic feet. 



62 SELF-INSTRUCTOR 

When a Board or Plank is broader at one End than the 
other, to find what Length of it will make a Foot, or 
any other required Quantity. 

Rule. — To the square of the product of the length and 
narrow end add twice the continual product of these quan- 
tities ; namely, the length, the difference between the breadths 
of the ends, and the area of the part required to be cut off. 
Extract the square root of the sum ; from the result deduct 
the product of the length and narrow end, and divide the 
remainder by the difference between the breadths of the 
ends. 

EXAMPLE. 

It is required to cut off 60 inches 
from the smaller end of a board ; 
A D being 3 inches, C E 6 inches, 
and A B 20 inches. 

Here A:r=^(y{( AB ><ADV + 4BCX 
ABX 60 j— A B X A D =i(v{( 20 XsV + 
6 X 20 X 60 | — 20 X 3 = 14.64, the length required. 

To find how much in Length will make a Solid Foot, or any 
other required Quantity, of Squared Timber, of equal 
Dimensions from End to End. 

Rule. — Divide 1,728 — the solid inches in a foot, or the 
solidity to be cut off — by the area of the end in inches. 

1. If a piece of timber be 14 inches broad and 10 inches 
deep, how much of it will make a solid foot ? 

Ans. 12i| inches, the length required. 

10 X H = 140 ; 1,728 -J- 140 = 12 Jf inches. 




ON LUMBER SURVEYING. 



63 



Side. — Multiply the area corresponding to the quarter 
girt in inches, by the length of the piece in feet, and the 

CtofT, ^ ^ f dUy - K the qUai ' tCT ^ exceeds the 
hunts of he table, take J of it, and 4 times the contents thus 
found will give the required contents. 

A Table for Measuring Timber. 



Quarter Girt. 


Area. 


Quarter Girt. 


Area. 


Quarter Girt. 


Area. 


Inches. 

6 

% 

7 


Feet. 
.250 


Inches. 

12 


Feet. 
1.000 


Inches. 

18 


Feet. 
2.250 


.272 
294 
.317 
.340 
.364 


12| 
12j 

12| 

13 

131 


1.042 
1.085 
1.129 
1.174 
1.219 


18i 

19 

19± 

20 

20i 

21 


2.376 
2.506 
2.640 
2.777 
! 2.917 


8 

r.1 


.390 


13j 


1.265 


3.062 


.417 


13j 


1.313 


2lJ 


3.209 


.444 
.472 


14 
14£ 


1.361 
1.410 


22 

22j 


3.362 
3.516 


8^ 

o3 


.501 


14 


1.460 


23 


3.673 


84 


.531 


14| 


1.511 


23j 


3.835 


9 


.562 


15 


1.562 


24 


4.000 


.594 


15i 

isi 


1.615 


24|- 


4.168 


^2 
«3 


.626 


1.668 


25 


4.340 


94 


.659 


15| 


1.722 


25j 


4.516 


10 


.694 


16 


1.777 


26 


4.694 


101 


.730 


161 


1833 


26^ \ 4.876 


10 i 


.766 


16i 1.890 


27 1 5.062 


lOf 


.803 


16| 1.948 


27^ J 5.252 


11 


.840 


17 2.006 


28 j 5.444 


nf 


.878 


171 


2.066 


28^ ' 5.640 


.918 


"i 


2.126 


29 


5.840 


.959 


17| 


2.187 


29j 


6.044 

1 



1 ^t? the C ° ntentS of a P iece of tim ber whose 
length is 30 feet and quarter girt is 17f inches. 

Ans. 65.610 feet. 



64 



SELF-INSTRUCTOR 



Solution by the Table. — • Look for the quarter girt 17 j, 
in the column marked Quarter Girt, and in the adjoining 
column marked Area, will be found 2.186, which multiplied 
by the length, 30 feet, will be 65.610 feet for the solid con- 
tents. 

Table showing the Weight in Pounds and Decimals of a 
Pound Avoirdupois of one Cubic Foot of the following 
Kinds of Wood. 



Cork Wood 15.00 | 

Poplar 23.94 

Larch or Hackmatack . . 34.00 

Elm and West India Fir . 34.75 

Mahogany 35.00 

Pitch Pine 41.25 

Cedar 37.25 

Pear Tree 41.31 

Walnut 41.94 

Elder Tree 43.44 

Beech 43.50 

Cherry Tree 44.68 



Maple and Kiga Fir . . 46.87 

Ash and Dantzic Oak . . 47. 50 

Apple Tree 49.56 

Alder 50.00 

Oak, Canadian .... 54.50 

Boxwood, French . . . 57. 00 

Logwood 57.06 

Oak, English 51.87 

Oak, sixty years old . . 73.12 

Ebony 83.18 

Lignum VitEe 83.31 



Rule for finding the Weight of any kind of Timber. 

Multiply the number of cubic feet it contains by the 
weight of one cubic foot of said timber. 



EXAMPLES. 

1. What is the weight of a piece of hackmatack timber 
8 inches X 12 inches, and 30 feet long? 

By the table given of cubic measure, 8 inches X 12 
inches is -§ of the length, for the contents ; therefore 30 -~- § 
= 20 feet, contents. 

By the table of weights a cubic foot of hackmatack is 
= to 34 lbs., therefore 34 X 30 = 1,020 lbs. avoirdupois. 



ON LUMBER SURVEYING. 65 

2. What is the weight of a piece of Canadian oak 12 
inches X 12 inches, and 30 feet long? Ans. 1,635.00 lbs. 

3. What is the weight of a piece of French boxwood 10 
inches X 12 inches, and 24 feet in length? 

By the table of cubic measure, 10 inches X 12 inches is 
f of the length, for the contents in cubic feet ; therefore 24 
-f- | == 20 feet, contents ; 20 X 57 = 1,140 lbs. = weight 
required. 

P. S. — The weight of any substance may be found as 
above, by finding the weight of 1 cubic foot and multiplying 
said weight by the contents. 

TONNAGE OF VESSELS. 

Government Rule. English, 

For vessels aground, the length is to be measured on a 
straight line along the rabbet of the keel, from a perpendicu- 
lar, let fall from the back of the main-post, at the height of 
the wing-transom, to a perpendicular at the height of the 
upper deck (but the middle deck of three-decked ships), 
from the forepart of the stern ; then from the length between 
these perpendiculars subtract three fifths of the extreme 
breadth for the rake of the stern ; and 2| inches for every 
foot of the height of the wing-transom above the lower part 
of the rabbet of the keel, for the rake abaft ; and the re- 
mainder will be the length of the keel for tonnage. The 
main breadth is to be taken from the outside of the outside 
plank, in the broadest part of the ship either above or be- 
low the wales, deducting therefrom all that it exceeds the 
thickness of the plank of the bottom, which shall be ac- 
counted the main breadth ; so that the moulding breadth, or 
the breadth of the frame, will then be less than the main 
breadth, so found, by double the thickness of the plank of 
the bottom. 

Rule. — Then multiply the length of the keel for tonnage, 
by the main breadth, so taken, and the product by half the 



66 SELF-INSTRUCTOR 

breadth ; then divide the whole by 94, and the quotient will 
be the tonnage. 

In cutters and brigs, where the rake of the stern-post ex- 
ceeds 2\ inches to every foot in height, the actual rake is 
generally subtracted instead of the 2\ inches to every foot, 
as before mentioned. 

1. Suppose the length from the fore-part of the stern, at 
the height of the upper deck, to the after-part of the stern- 
post, at the height of the wing-transom to be 115 feet 8 
inches, the breadth from outside to outside 40 feet 6 inches, 
and the height of the wing-transom 21 feet 10 inches, what is 
the tonnage ? Ans. 1,094. 

ft. in. 

40 6 breadth 
3 

40~3"X3= 120.9 ; 120.9 -f5= 24.15. 

21.10 height of wing-transom 21.10 X 2^ = 54^ ; 54 T 7 ^ 
-=- 12 = 4.55 ; 4.55 -f- 24.15 = 28.70 ; 155.66 — 28.70 === 
126.96= length. 

126.96X40.25X20 ^25 = lfiUj ^ t(Hmage required . 

2. If • the length of the keel be 120 feet, and the breadth 
40 feet, what is the tonnage? Arts. 1,021^ tons. 

Solution. — 120 X 40 = 4,800; 4,800 X 20= 96,000 ; 
96,000 -^ 94 = l,021if tons. 

3. If the length of the keel be 80 feet, and the breadth 
of the beam 36 feet, what is the tonnage ? Arts. 55 Iff. 

4. If the length of the keel be 460 feet, and the breadth 
of the beam 80 feet, what is the tonnage. 

Ans. 15,659 tons. 
Some divide the last product by 100, to find the ton- 
nage of king's ships, and by 95, to find that of merchant 
ships. 

American Government Rule. 

For single-decked vessels. — Take the length on deck from 
the forward side of the main stern to the after-side of the 
stern-post, and the breadth at the broadest part above the 



ON LUMBER SURVEYING. 67 

main wales ; take the depth from the under side of the deck 
plank to the ceiling of the hold, and deduct from the length 
three fifths of the breadth ; multiply the remainder by the 
breadth, and the product by the depth, and divide the last 
product by 95. 

For double-decked vessels. — Proceed as with single-decked 
vessels, except for the depth take half the breadth. 

GAUGING. 

Gauging signifies the art of measuring all kinds of vessels 
and determining their capacity or the quantity of fluid or 
other matter they contain. It is usual to divide casks into 
four varieties, which are judged of from the greater or less 
apparent curvature of their sides, namely : — 

1. The middle frustum of a spheroid. 

2. The middle frustum of a parabolic spindle. 

3. T^e two equal frustums of a paraboloid. 

4. The two equal frustums of a cone. 

282 cubic inches make 1 ale gallon, or beer. 
231 cubic inches make 1 wine gallon. 
21,504 cubic inches make 1 malt bushel. 

To find the contents of a Cash by the Mean Diameter. 

Ride. — Multiply the difference of the head and bung di- 
ameters by .68 for the first variety ; by .62 for the second ; 
by .55 for the third ; and by .5 for the fourth, when the dif- 
ference between the head* and bung diameter is less than 6 
inches; but when the difference between these exceeds 6 
inches, multiply that difference by .7 for the first variety ; by 
.64 for the second; by .57 for the third; and by .52 for the 
fourth. Add this product to the head diameter, and the 
sum will be a mean diameter. Square this mean diameter, 
and multiply the square by the length of the cask ; this 
product multiplied or divided by the proper multiplier or 
divisor, will give the contents. 

1. What are the contents of a spheroidal cask, whose 



68 SELF-INSTRUCTOK 

length is 40 inches, bung diameter 32 inches, and head di- 
ameter 24 inches ? Ans. 97.6 gallons. 
Solutio?i.-32 - 24= 8 ; 8 X 7 = 5.6 ; 5.6 + 24= 29.6 
= mean diameter; 29.6 X 29.6 = 876.1 6 = square ; 876.16 
X 40 = 35046.40, which being divided by 359.5, the divisor 
for imperial gallons, will be equal to 97.6 gallons. 
By the gauging rule — 

Set 40 on C. to the G. E. 18.79 on D. against 
24 on D. stands 64.99 on C. 

32 on D. stands 116.2 on C. 

+ 116.2 



3)297.39 



99.13 gallons. 



& i 



Dr. HuttorCs General Rule for finding the Contents of Casks. 

Add into one sum 39 times the square of the bung diame- 
ter, 25 times the square of the head diameter, and 26 times 
the product of the two diameters ; then multiply the sum by 
the length, and the product again by .0003 If for the con- 
tents in gallons. 

EXAMPLE. 

1. What are the contents of a cask whose length is 40 
inches, and the bung and head diameters 32 and 24 ? 

Ans. 93.4579 gallons. 
32 X 32 = 1024 ; 1024 X 39 = 39936 
24X24= 576; 576X25 = 14400 
32X24= 768; 768X26 = 19968 



74304 X 40 = 2972160 



.0003 If 



93.4579 
Ullaging is the art of finding what quantity of liquor is 
contained in a cask when partly empty. And it is consid- 



ON LUMBER SURVEYING. 69 

ered in two positions ; first, as standing on its end ; secondly, 
lying on its side. 

To find the Contents of Ullage by the Sliding Rule. 
By one of the preceding problems find the whole con- 
tents of the cask. Then set the length on N. to 100 on S. 
S. for a segment standing, or set the bung diameter on N. to 
100 on S. L. for a segment lying; then against the wet 
inches on N. is a number on S. S. or S. L. to be reserved. 
Next set 100 on B. to the reserved number on A. ; then 
against the whole contents on B. will be found the ullage 
on A. 

QUESTIONS FOR EXERCISE. 

1. What are the contents of 20 pieces of timber 8 inches 
X 12 inches, and 36 feet long in cubic feet, and also in su- 
perficial feet ? 

2. What number of cubic feet in a log whose quarter girt 
is 17^ inches and length 18 feet ? 

3. What are the contents of 24 logs 16 feet long whose 
quarter girt is 27 inches ? 

4. Required the tonnage of a ship by the English and 
American rules, the length of the keel being 125 feet and 
the breadth of the beam 42 feet ? 

5. What is the weight of a piece of hackmatack timber 8 
inches X 10 inches and 28 feet in length ? 

6. Required the number of tons in 16 pieces of timber 
24 feet long and 12 inches X 16 inches? 

7. In 2,500 feet running length of 2 inches X 10 inches, 
how many feet of board measure ? 

8. In 300 feet running length of 10 inch X 12 inch tim- 
ber, how many tons ? 

9. What are the contents of a cask of the first variety in 
wine and ale gallons, whose length is 50 inches, bung diam- 
eter 38 inches, and head diameter 30 inches ? 

10. If a log be 35 inches in diameter, what is the largest 
piece of square timber that can be sawed from it ? 



70 SELF-INSTRUCTOR 

11. What difference is there between a floor 28 feet long 
X 20 feet broad, and two others, each of half the dimen- 
sions ; and what do the three floors come to @ $9.00 per 
100 square feet ? Ans. $75.-60. 

12. An elm plank is 14 feet 3 inches long, and it is de- 
sired that just a square yard may be slit off from it ; at what 
distance from the edge must the line be struck ? 

Ans. 7 t 9 t? t inches. 
18. A joist is 7 inches wide and 2^ inches thick, but a 
scantling just as big again, that shall be 3 inches thick, is 
wanted ; what will the other dimension be ? 

Ans. 1 If inches. 

14. The perambulator is so contrived as to turn just 
twice in 16J feet ; required the diameter ? Ans. 2.626 feet. 

15. In turning a chaise within a ring of a certain diameter, 
it was observed that the outer wheel made two revolutions 
while the inner made but one ; the wheels were both 4 feet 
high, and supposing them fixed at the distance of 5 feet 
asunder on the axletree, what was the circumference of the 
track described by the outside wheel ? Ans. 63 feet nearly. 

16. Having a rectangular board 58 inches by 27 inches, 
I would have a square foot cut off parallel to the shorter 
edge ; I would then have the same quantity cut from the 
remainder, parallel to the longer, and this alternately re- 
peated, till there shall not be the quantity of a foot left ; 
what will be the dimensions of the remaining piece ? 

Ans. 20.7 inches by 6.086. 

17. What is the length of a chord which cuts off ^ of 
the area of a circle, whose diameter is 289 ? 

Ans. 278.6716. 

18. What will the diameter of a globe be, when the solid- 
ity and superficial contents are expressed by the same num- 
ber? Ans. 6. 

19. A gentleman has a garden 100 feet long and 80 feet 
broad, and a gravel walk is to be made of an equal width 
half round it ; what must be the breadth of the walk to 
take up just half the ground? Ans 25.968 feet. 



ON LUMBER SURVEYING. 71 

20. How many 3-inch cubes may be cut out of a 12-inch 
cube? Ans. 64. 

21. How high above the earth must a person be raised 
that he may see one third of its surface ? 

Ans. To the height of the earth's diameter. 

22. How many feet of boards would cover the surface of 
the earth, its diameter being 7,958 miles ; and how many 
solid feet in it ? 

f 5,546,407,680,000,000. No. of 
feet of boards to cover it. 
17,416,291,092,323,844,085,000. 
No. of cubic feet in the earth. 

23. If the diameter of a circle be 50 feet, what is the 
circumference of it ? 

24. Two pillars standing on a horizontal plane are 120 
feet asunder ; the height of the higher is 100 feet, and that 
of the lower 80 ; whereabout in the plane must a person 
place himself, so that his distance from the top of either of 
the pillars shall be equal to the distance between them ? 

Ans. 91.78 feet from the bottom of the lower. 
69.92 feet from the bottom of the other. 

25. Three ships are equally distant from an island, the 
first ship is 30 miles from the second, the second is 25 miles 
from the third, and the third is 20 miles from the first ; re- 
quired the distance to the isle ? 

Ans. 15.118579 miles from each. 

26. Prove that the elevation of the North or Polar star 
above the horizon is equal to the latitude of the place where 
its altitude is taken. 

27. I have a board in the form of a triangle ; the length 
of one of its sides is 1 6 feet. I wish to sell one half of it ; 
at what distance from the larger end must it be divided par- 
allel to the larger end. Ans. 4.68 feet. 

28. In 2,500 feet running lengths of 7 inches X 9 inches, 
how many feet running lengths of 2 J inches X H ? 



72 



SELF-INSTRUCTOR 



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ON LUMBEK SURVEYING. 



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ON LUMBEE SUEVEYING. 77 

How to use the Log or Timber Rule. 

If the timber is tapering, the girt should be taken about 
one third the distance from the larger to the smaller end. 
Some take the girt in the middle. Girt the log to be meas- 
ured, and take the quarter of it, and measure the length of 
the log. Then look along the top of the table till you come 
to the corresponding quarter girt ; then run down the column 
underneath the quarter girt till you get opposite the length, - 
where you will find the contents. Or, you can find the con- 
tents by taking the diameter of the small end and the length. 
Then find the corresponding diameter at the foot of the table, 
and ascend the line perpendicularly till you come opposite 
the length, where you will find the contents. 

P. S. — This table allows one fourth of the true contents 
of the log for bark, saw kerf, and waste slab. It has been 
extensively used by timber merchants, and is just about as 
fair a rule to go by as any I have seen. There are many 
allowances to be made which are left to the scaler's judg- 
ment, and for which it would be almost impossible to make 
due allowance in the table. 



INTEREST. 



Rule for finding the interest at 6 per cent. — Multiply the 
sum by the number of days, divide the product by 6, then 
strike off the right-hand figure. 

EXAMPLE. 

$200 
12 days. 



6)2400 



400 = 40 cents is the interest. 



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Ans. 82 cents. 



INTEREST. 



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Interest on $50 for 3 years = $9.00 
Interest on $50 for 2 mos. = 50 
Interest on $50 for 10 days = 8 



Arts. $9.58 



WANTED. 

Agents to sell this Book, throughout the United States, 
the Dominion of Canada, California, and Oregon. Exclu- 
sive territory given. Good inducements to agents. The 
book will be sent to any address, free of postage, on receipt 
of Two Dollars. Send Post-Office orders, or by express. 
Address 

CHARLES KINSLEY, 
Calais, Me., or St. Stephen, N. B. 



NOTE. 

All Lumber Manufacturers, Lumber Dealers, Millmen, 
Carpenters, Carriage Makers, Shipbuilders, Cabinet Mak- 
ers, Ship Brokers, Ship Carpenters, Railroad Conductors, 
Engineers, Machinists, Freight Agents, Teachers, Students, 
Architects, Merchants, Accountants, and others, will find 
it to their advantage to procure a copy of this book, as the 
knowledge it imparts may save them in a few years' prac- 
tice hundreds of dollars. The book contains twelve new 
rules for finding the superficial contents of lumber, which 
do the same work as one hundred and fifty of the rules 
generally used. 

CHARLES KINSLEY. 






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